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y2+13y-5=0
We add all the numbers together, and all the variables
y^2+13y-5=0
a = 1; b = 13; c = -5;
Δ = b2-4ac
Δ = 132-4·1·(-5)
Δ = 189
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}$$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{189}=\sqrt{9*21}=\sqrt{9}*\sqrt{21}=3\sqrt{21}$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3\sqrt{21}}{2*1}=\frac{-13-3\sqrt{21}}{2} $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3\sqrt{21}}{2*1}=\frac{-13+3\sqrt{21}}{2} $
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