13=-2(y-4)*3y

Solution for 13=-2(y-4)*3y equation:

13=-2(y-4)*3y

We move all terms to the left:

13-(-2(y-4)*3y)=0


We calculate terms in parentheses: -(-2(y-4)*3y), so:

-2(y-4)*3y

We multiply parentheses

-6y^2+24y

Back to the equation:

-(-6y^2+24y)


We get rid of parentheses

6y^2-24y+13=0

a = 6; b = -24; c = +13;
Δ = b2-4ac
Δ = -242-4·6·13
Δ = 264
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:

$\sqrt{\Delta}=\sqrt{264}=\sqrt{4*66}=\sqrt{4}*\sqrt{66}=2\sqrt{66}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-2\sqrt{66}}{2*6}=\frac{24-2\sqrt{66}}{12}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+2\sqrt{66}}{2*6}=\frac{24+2\sqrt{66}}{12}
$