20^=y(y+30)

Solution for 20^=y(y+30) equation:

20^=y(y+30)

We move all terms to the left:

20^-(y(y+30))=0

We add all the numbers together, and all the variables

-(y(y+30))=0


We calculate terms in parentheses: -(y(y+30)), so:

y(y+30)

We multiply parentheses

y^2+30y

Back to the equation:

-(y^2+30y)


We get rid of parentheses

-y^2-30y=0

We add all the numbers together, and all the variables

-1y^2-30y=0

a = -1; b = -30; c = 0;
Δ = b2-4ac
Δ = -302-4·(-1)·0
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-30}{2*-1}=\frac{0}{-2}
=0
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+30}{2*-1}=\frac{60}{-2}
=-30
$