-16x^2+40x=-14

Solution for -16x^2+40x=-14 equation:

-16x^2+40x=-14

We move all terms to the left:

-16x^2+40x-(-14)=0

We add all the numbers together, and all the variables

-16x^2+40x+14=0

a = -16; b = 40; c = +14;
Δ = b2-4ac
Δ = 402-4·(-16)·14
Δ = 2496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2496}=\sqrt{64*39}=\sqrt{64}*\sqrt{39}=8\sqrt{39}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-8\sqrt{39}}{2*-16}=\frac{-40-8\sqrt{39}}{-32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+8\sqrt{39}}{2*-16}=\frac{-40+8\sqrt{39}}{-32}
$