(-7+2y)(-7+2y)+4y^2=25

Solution for (-7+2y)(-7+2y)+4y^2=25 equation:

(-7+2y)(-7+2y)+4y^2=25

We move all terms to the left:

(-7+2y)(-7+2y)+4y^2-(25)=0

We add all the numbers together, and all the variables

4y^2+(2y-7)(2y-7)-25=0

We multiply parentheses ..

4y^2+(+4y^2-14y-14y+49)-25=0

We get rid of parentheses

4y^2+4y^2-14y-14y+49-25=0

We add all the numbers together, and all the variables

8y^2-28y+24=0

a = 8; b = -28; c = +24;
Δ = b2-4ac
Δ = -282-4·8·24
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4}{2*8}=\frac{24}{16}
=1+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4}{2*8}=\frac{32}{16}
=2
$