(t)=-8t2+32t+40

Solution for (t)=-8t2+32t+40 equation:

(t)=-8t^2+32t+40

We move all terms to the left:

(t)-(-8t^2+32t+40)=0

We get rid of parentheses

8t^2-32t+t-40=0

We add all the numbers together, and all the variables

8t^2-31t-40=0

a = 8; b = -31; c = -40;
Δ = b2-4ac
Δ = -312-4·8·(-40)
Δ = 2241
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2241}=\sqrt{9*249}=\sqrt{9}*\sqrt{249}=3\sqrt{249}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-3\sqrt{249}}{2*8}=\frac{31-3\sqrt{249}}{16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+3\sqrt{249}}{2*8}=\frac{31+3\sqrt{249}}{16}
$