16t^2+36t=-10

Solution for 16t^2+36t=-10 equation:

16t^2+36t=-10

We move all terms to the left:

16t^2+36t-(-10)=0

We add all the numbers together, and all the variables

16t^2+36t+10=0

a = 16; b = 36; c = +10;
Δ = b2-4ac
Δ = 362-4·16·10
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{41}}{2*16}=\frac{-36-4\sqrt{41}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{41}}{2*16}=\frac{-36+4\sqrt{41}}{32}
$