Question

6x²-7x=20 solve the following quadratic equation

Answer 1

Answer:

$x1 = - \frac{4}{3}$

$x2 = \frac{5}{2}$

Step-by-step explanation:

$6 {x}^{2} - 7x = 20$

Move constant to the left and change its sign

${6x}^{2} - 7x - 20 = 0$

Write -7x as a difference

$6 {x}^{2} + 8x - 15x - 20 = 0$

Factor out 2x from the expression

$2x(3x + 4) - 15x - 20 = 0$

Factor out -5 from the expression

$2x(3x + 4) - 5(3x + 4) = 0$

Factor out 3x + 4 from the expression

$(3x + 4)(2x - 5) = 0$

When the product of factors equals 0 , at least one factor is 0

$3x + 4 = 0$

$2x - 5 = 0$

Solve the equation for X1

$3x + 4 = 0$

Move constant to right side and change its sign

$3x = 0 - 4$

Calculate the difference

$3x = - 4$

Divide both sides of the equation by 3

$\frac{3x}{3} = \frac{ - 4}{3}$

Calculate

$x = - \frac{4}{3}$

Again,

Solve for x2

$2x - 5 = 0$

Move constant to right side and change its sign

$2x = 0 + 5$

Calculate the sum

$2x = 5$

Divide both sides of the equation by 2

$\frac{2x}{2} = \frac{5}{2}$

Calculate

$x = \frac{5}{2}$

$x1 = - \frac{4}{3}$

$x2 = \frac{5}{2}$

Hope this helps...

Best regards!!

Answer 2

Answer:

x = -4/3 and x = 5/2.

Step-by-step explanation:

6x² - 7x = 20

6x² - 7x - 20 = 0

To solve this, we can use the quadratic formula to solve this.

[please ignore the A-hat; that is a bug]

$\frac{-b±\sqrt{b^2 - 4ac} }{2a}$

In this case, a = 6, b = -7, and c = -20.

$\frac{-(-7)±\sqrt{(-7)^2 - 4 * 6 * (-20)} }{2(6)}$

= $\frac{7±\sqrt{49 + 80 * 6} }{12}$

= $\frac{7±\sqrt{49 + 480} }{12}$

= $\frac{7±\sqrt{529} }{12}$

= $\frac{7±23 }{12}$

$\frac{7 - 23 }{12}$ = $\frac{-16 }{12}$ = -8 / 6 = -4 / 3

$\frac{7 + 23 }{12}$ = $\frac{30}{12}$ = 15 / 6 = 5 / 2

So, x = -4/3 and x = 5/2.

Hope this helps!