7y2+13=41

Solution for 7y2+13=41 equation:

7y^2+13=41

We move all terms to the left:

7y^2+13-(41)=0

We add all the numbers together, and all the variables

7y^2-28=0

a = 7; b = 0; c = -28;
Δ = b2-4ac
Δ = 02-4·7·(-28)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28}{2*7}=\frac{-28}{14}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28}{2*7}=\frac{28}{14}
=2
$