20=4y(y-13)

Solution for 20=4y(y-13) equation:

20=4y(y-13)

We move all terms to the left:

20-(4y(y-13))=0


We calculate terms in parentheses: -(4y(y-13)), so:

4y(y-13)

We multiply parentheses

4y^2-52y

Back to the equation:

-(4y^2-52y)


We get rid of parentheses

-4y^2+52y+20=0

a = -4; b = 52; c = +20;
Δ = b2-4ac
Δ = 522-4·(-4)·20
Δ = 3024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3024}=\sqrt{144*21}=\sqrt{144}*\sqrt{21}=12\sqrt{21}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-12\sqrt{21}}{2*-4}=\frac{-52-12\sqrt{21}}{-8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+12\sqrt{21}}{2*-4}=\frac{-52+12\sqrt{21}}{-8}
$