4v^2=25v+21

Solution for 4v^2=25v+21 equation:

4v^2=25v+21

We move all terms to the left:

4v^2-(25v+21)=0

We get rid of parentheses

4v^2-25v-21=0

a = 4; b = -25; c = -21;
Δ = b2-4ac
Δ = -252-4·4·(-21)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-31}{2*4}=\frac{-6}{8}
=-3/4
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+31}{2*4}=\frac{56}{8}
=7
$