t=-2t^2+32t+6

Solution for t=-2t^2+32t+6 equation:

t=-2t^2+32t+6

We move all terms to the left:

t-(-2t^2+32t+6)=0

We get rid of parentheses

2t^2-32t+t-6=0

We add all the numbers together, and all the variables

2t^2-31t-6=0

a = 2; b = -31; c = -6;
Δ = b2-4ac
Δ = -312-4·2·(-6)
Δ = 1009
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-\sqrt{1009}}{2*2}=\frac{31-\sqrt{1009}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{1009}}{2*2}=\frac{31+\sqrt{1009}}{4}
$