-7v^2+24v=16

Solution for -7v^2+24v=16 equation:

-7v^2+24v=16

We move all terms to the left:

-7v^2+24v-(16)=0

a = -7; b = 24; c = -16;
Δ = b2-4ac
Δ = 242-4·(-7)·(-16)
Δ = 128
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{2}}{2*-7}=\frac{-24-8\sqrt{2}}{-14}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{2}}{2*-7}=\frac{-24+8\sqrt{2}}{-14}
$