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y2+4y-7=5
We move all terms to the left:
y2+4y-7-(5)=0
We add all the numbers together, and all the variables
y^2+4y-12=0
a = 1; b = 4; c = -12;
Δ = b2-4ac
Δ = 42-4·1·(-12)
Δ = 64
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{64}=8$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*1}=\frac{-12}{2} =-6 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*1}=\frac{4}{2} =2 $
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