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x^2+3x-10=2x^2-3x-2
We move all terms to the left:
x^2+3x-10-(2x^2-3x-2)=0
We get rid of parentheses
x^2-2x^2+3x+3x+2-10=0
We add all the numbers together, and all the variables
-1x^2+6x-8=0
a = -1; b = 6; c = -8;
Δ = b2-4ac
Δ = 62-4·(-1)·(-8)
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2}{2*-1}=\frac{-8}{-2} =+4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2}{2*-1}=\frac{-4}{-2} =+2 $
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