7v+4v^2=21

Solution for 7v+4v^2=21 equation:

7v+4v^2=21

We move all terms to the left:

7v+4v^2-(21)=0

a = 4; b = 7; c = -21;
Δ = b2-4ac
Δ = 72-4·4·(-21)
Δ = 385
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{385}}{2*4}=\frac{-7-\sqrt{385}}{8}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{385}}{2*4}=\frac{-7+\sqrt{385}}{8}
$