4y^2=5(3y+5)

Solution for 4y^2=5(3y+5) equation:

4y^2=5(3y+5)

We move all terms to the left:

4y^2-(5(3y+5))=0


We calculate terms in parentheses: -(5(3y+5)), so:

5(3y+5)

We multiply parentheses

15y+25

Back to the equation:

-(15y+25)


We get rid of parentheses

4y^2-15y-25=0

a = 4; b = -15; c = -25;
Δ = b2-4ac
Δ = -152-4·4·(-25)
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{-(-15)-25}{2*4}=\frac{-10}{8}
=-1+1/4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+25}{2*4}=\frac{40}{8}
=5
$