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y2+2y=48
We move all terms to the left:
y2+2y-(48)=0
We add all the numbers together, and all the variables
y^2+2y-48=0
a = 1; b = 2; c = -48;
Δ = b2-4ac
Δ = 22-4·1·(-48)
Δ = 196
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{196}=14$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-14}{2*1}=\frac{-16}{2} =-8 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+14}{2*1}=\frac{12}{2} =6 $
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