y2+14y=13

Solution for y2+14y=13 equation:

y2+14y=13

We move all terms to the left:

y2+14y-(13)=0

We add all the numbers together, and all the variables

y^2+14y-13=0

a = 1; b = 14; c = -13;
Δ = b2-4ac
Δ = 142-4·1·(-13)
Δ = 248
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{248}=\sqrt{4*62}=\sqrt{4}*\sqrt{62}=2\sqrt{62}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{62}}{2*1}=\frac{-14-2\sqrt{62}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{62}}{2*1}=\frac{-14+2\sqrt{62}}{2}
$