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2x^2+3x+2=x2+2x+8
We move all terms to the left:
2x^2+3x+2-(x2+2x+8)=0
We add all the numbers together, and all the variables
2x^2-(+x^2+2x+8)+3x+2=0
We get rid of parentheses
2x^2-x^2-2x+3x-8+2=0
We add all the numbers together, and all the variables
x^2+x-6=0
a = 1; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·1·(-6)
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*1}=\frac{-6}{2} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*1}=\frac{4}{2} =2 $
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