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x^2-4x=2x^2+5x+10
We move all terms to the left:
x^2-4x-(2x^2+5x+10)=0
We get rid of parentheses
x^2-2x^2-4x-5x-10=0
We add all the numbers together, and all the variables
-1x^2-9x-10=0
a = -1; b = -9; c = -10;
Δ = b2-4ac
Δ = -92-4·(-1)·(-10)
Δ = 41
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{41}}{2*-1}=\frac{9-\sqrt{41}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{41}}{2*-1}=\frac{9+\sqrt{41}}{-2} $
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