y^2+14y-51=0

Solution for y^2+14y-51=0 equation:

y^2+14y-51=0

a = 1; b = 14; c = -51;
Δ = b2-4ac
Δ = 142-4·1·(-51)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-20}{2*1}=\frac{-34}{2}
=-17
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+20}{2*1}=\frac{6}{2}
=3
$