(4+1)^2=2y^2-y-3

Solution for (4+1)^2=2y^2-y-3 equation:

(4+1)^2=2y^2-y-3

We move all terms to the left:

(4+1)^2-(2y^2-y-3)=0

We add all the numbers together, and all the variables

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

We add all the numbers together, and all the variables

-(2y^2-y-3)+25=0

We get rid of parentheses

-2y^2+y+3+25=0

We add all the numbers together, and all the variables

-2y^2+y+28=0

a = -2; b = 1; c = +28;
Δ = b2-4ac
Δ = 12-4·(-2)·28
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-15}{2*-2}=\frac{-16}{-4}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+15}{2*-2}=\frac{14}{-4}
=-3+1/2
$