Y=2x^2+5(3x+2x)

Solution for Y=2x^2+5(3x+2x) equation:

=2Y^2+5(3Y+2Y)

We move all terms to the left:

-(2Y^2+5(3Y+2Y))=0

We add all the numbers together, and all the variables

-(2Y^2+5(+5Y))=0


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

2Y^2+5(+5Y)

We multiply parentheses

2Y^2+25Y

Back to the equation:

-(2Y^2+25Y)


We get rid of parentheses

-2Y^2-25Y=0

a = -2; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·(-2)·0
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*-2}=\frac{0}{-4}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*-2}=\frac{50}{-4}
=-12+1/2
$