4n^2-20=7n

Solution for 4n^2-20=7n equation:

4n^2-20=7n

We move all terms to the left:

4n^2-20-(7n)=0

a = 4; b = -7; c = -20;
Δ = b2-4ac
Δ = -72-4·4·(-20)
Δ = 369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3\sqrt{41}}{2*4}=\frac{7-3\sqrt{41}}{8}
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3\sqrt{41}}{2*4}=\frac{7+3\sqrt{41}}{8}
$