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x^2-2x-5=-4x+10
We move all terms to the left:
x^2-2x-5-(-4x+10)=0
We get rid of parentheses
x^2-2x+4x-10-5=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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