Question

H (t) = -t2 + t + 12 Solve by using quadratic formula

Answer 1

The quadratic formula is:

$x=\frac{-b+\sqrt{(b)^2-4(a)(c)}}{2a}$

and

$x=\frac{-b-\sqrt{(b)^2-4(a)(c)}}{2a}$

Let's identify our a, b, and c values:

a: -1

b: 1

c: 12

Plug in the values for a, b, and c into the equation. Let's do the first equation:

$x=\frac{-1+\sqrt{(1)^2-4(-1)(12)}}{2(-1)}$

Simplify everything in the radical:

$x=\frac{-1+\sqrt{49}}{-2}$

Simplify the radical:

$x=\frac{-1+7}{-2}$

Combine like terms:

$x=\frac{6}{-2}$

Simplify:

$x=-3$

This is one solution, now, let's solve for the other equation:

Since when simplified, everything is the same except the subtraction sign, we can skip the simplification again and change the sign to subtraction:

$x=\frac{-1-7}{-2}$

Combine like terms:

$x=\frac{-8}{-2}$

Simplify:

$x=4$

Your final answers are:

$x=-3$

$x=4$

Answer 2

$ax^2+bx+c=0$

$H(t)=-t^2+t+12$

a = -1; b = +1; c = 12

Quadratic Formula: $\frac{-b±\sqrt{b^2-4ac} }{2a} \\ \\ \frac{-(1)±\sqrt{(1)^2-4(-1)(12)} }{2(-1)} \\ \\ \frac{-1±\sqrt{(1)+4(12)} }{-2}\\ \\ \frac{-1±\sqrt{(1)+48} }{-2}\\ \\ \frac{-1±\sqrt{49} }{-2}\\ \\ \frac{-1±7 }{-2}\\ \\ \frac{-1+7 }{-2}...and...\frac{-1-7 }{-2}\\ \\ -3...and...+4$

This should be your final answer. So sorry about the capital A if it's in the whole work I did here. I can't get rid of it whenever I put plus and minus sign. I remember saying this and they had a solution for it on Brainly but I forgot what that was...