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(x^2+7x-9)-(2x^2-3x+4)=x
We move all terms to the left:
(x^2+7x-9)-(2x^2-3x+4)-(x)=0
We add all the numbers together, and all the variables
-1x+(x^2+7x-9)-(2x^2-3x+4)=0
We get rid of parentheses
x^2-2x^2-1x+7x+3x-9-4=0
We add all the numbers together, and all the variables
-1x^2+9x-13=0
a = -1; b = 9; c = -13;
Δ = b2-4ac
Δ = 92-4·(-1)·(-13)
Δ = 29
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{29}}{2*-1}=\frac{-9-\sqrt{29}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{29}}{2*-1}=\frac{-9+\sqrt{29}}{-2} $
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