G=-t^2+8t+10

Solution for G=-t^2+8t+10 equation:

=-G^2+8G+10

We move all terms to the left:

-(-G^2+8G+10)=0

We get rid of parentheses

G^2-8G-10=0

a = 1; b = -8; c = -10;
Δ = b2-4ac
Δ = -82-4·1·(-10)
Δ = 104
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$G_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{26}}{2*1}=\frac{8-2\sqrt{26}}{2}
$
$G_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{26}}{2*1}=\frac{8+2\sqrt{26}}{2}
$