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x2+12x-13=10x+2
We move all terms to the left:
x2+12x-13-(10x+2)=0
We add all the numbers together, and all the variables
x^2+12x-(10x+2)-13=0
We get rid of parentheses
x^2+12x-10x-2-13=0
We add all the numbers together, and all the variables
x^2+2x-15=0
a = 1; b = 2; c = -15;
Δ = b2-4ac
Δ = 22-4·1·(-15)
Δ = 64
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*1}=\frac{-10}{2} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*1}=\frac{6}{2} =3 $
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