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y2-3y-4=0
We add all the numbers together, and all the variables
y^2-3y-4=0
a = 1; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·1·(-4)
Δ = 25
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{25}=5$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-5}{2*1}=\frac{-2}{2} =-1 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+5}{2*1}=\frac{8}{2} =4 $
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