Answer:
Yes. Her solution is correct.
Step-by-step explanation:
Let's check if Jenna solution is correct:
To solve the equation 2x^2 +5x - 42 = 0, we can use Bhaskara's formula:
D = b^2 - 4ac = 25 + 4*2*42 = 25+336 = 361
sqrt(D) = 19
x1 = (-5 + 19)/4 = 14/4 = 7/2
x2 = (-5 - 19)/4 = -24/4 = -6
We must agree with Jenna's solution, because the values she found as solution are correct: with we replace these values of x in the equation, we will find 0 = 0, which is correct and proves that these values are the solution of the equation.
Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.
![~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20distance%20between%202%20points%7D%7D%7Bd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%7D~%5Chfill~%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B10%7D~%2C~%5Cstackrel%7By_1%7D%7B5%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B15%7D~%2C~%5Cstackrel%7By_2%7D%7B15%7D%29%20~%5Chfill%20a%3D%5Csqrt%7B%5B%2015-%2010%5D%5E2%20%2B%20%5B%2015-%205%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7Ba%3D%5Csqrt%7B125%7D%7D%20%5C%5C%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B15%7D~%2C~%5Cstackrel%7By_1%7D%7B15%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B30%7D~%2C~%5Cstackrel%7By_2%7D%7B9%7D%29%20~%5Chfill%20b%3D%5Csqrt%7B%5B%2030-%2015%5D%5E2%20%2B%20%5B%209-%2015%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7Bb%3D%5Csqrt%7B261%7D%7D)
![(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B30%7D~%2C~%5Cstackrel%7By_1%7D%7B9%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B10%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%20~%5Chfill%20c%3D%5Csqrt%7B%5B%2010-%2030%5D%5E2%20%2B%20%5B%205-%209%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20~%5Chfill%20%5Cboxed%7Bc%3D%5Csqrt%7B416%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}](https://tex.z-dn.net/?f=%5Cqquad%20%5Ctextit%7BHeron%27s%20area%20formula%7D%20%5C%5C%5C%5C%20A%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D%5Cqquad%20%5Cbegin%7Bcases%7D%20s%3D%5Cfrac%7Ba%2Bb%2Bc%7D%7B2%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D%5Csqrt%7B125%7D%5C%5C%20b%3D%5Csqrt%7B261%7D%5C%5C%20c%3D%5Csqrt%7B416%7D%5C%5C%20s%5Capprox%2023.87%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20A%5Capprox%5Csqrt%7B23.87%2823.87-%5Csqrt%7B125%7D%29%2823.87-%5Csqrt%7B261%7D%29%2823.87-%5Csqrt%7B416%7D%29%7D%5Cimplies%20%5Cboxed%7BA%5Capprox%2090%7D)
<h2>I Think it might be this hope you are helped by this</h2><h2></h2><h2>10(5x−6)</h2>
Answer:
1 hour and 45 minutes
Hope this helped, have a nice day! :)
