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(2x-3)+(2+x^2)=23
We move all terms to the left:
(2x-3)+(2+x^2)-(23)=0
We get rid of parentheses
x^2+2x+2-3-23=0
We add all the numbers together, and all the variables
x^2+2x-24=0
a = 1; b = 2; c = -24;
Δ = b2-4ac
Δ = 22-4·1·(-24)
Δ = 100
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{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*1}=\frac{-12}{2} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*1}=\frac{8}{2} =4 $
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