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x^2+10x-13=7
We move all terms to the left:
x^2+10x-13-(7)=0
We add all the numbers together, and all the variables
x^2+10x-20=0
a = 1; b = 10; c = -20;
Δ = b2-4ac
Δ = 102-4·1·(-20)
Δ = 180
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{5}}{2*1}=\frac{-10-6\sqrt{5}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{5}}{2*1}=\frac{-10+6\sqrt{5}}{2} $
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