Y=2x(4x-5)

Solution for Y=2x(4x-5) equation:

=2Y(4Y-5)

We move all terms to the left:

-(2Y(4Y-5))=0


We calculate terms in parentheses: -(2Y(4Y-5)), so:

2Y(4Y-5)

We multiply parentheses

8Y^2-10Y

Back to the equation:

-(8Y^2-10Y)


We get rid of parentheses

-8Y^2+10Y=0

a = -8; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-8)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-8}=\frac{-20}{-16}
=1+1/4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-8}=\frac{0}{-16}
=0
$