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y2+15y+56=0
We add all the numbers together, and all the variables
y^2+15y+56=0
a = 1; b = 15; c = +56;
Δ = b2-4ac
Δ = 152-4·1·56
Δ = 1
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{1}=1$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-1}{2*1}=\frac{-16}{2} =-8 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+1}{2*1}=\frac{-14}{2} =-7 $
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