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x^2+2x-5=9
We move all terms to the left:
x^2+2x-5-(9)=0
We add all the numbers together, and all the variables
x^2+2x-14=0
a = 1; b = 2; c = -14;
Δ = b2-4ac
Δ = 22-4·1·(-14)
Δ = 60
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{15}}{2*1}=\frac{-2-2\sqrt{15}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{15}}{2*1}=\frac{-2+2\sqrt{15}}{2} $
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