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3x+9=2x^2-2
We move all terms to the left:
3x+9-(2x^2-2)=0
We get rid of parentheses
-2x^2+3x+2+9=0
We add all the numbers together, and all the variables
-2x^2+3x+11=0
a = -2; b = 3; c = +11;
Δ = b2-4ac
Δ = 32-4·(-2)·11
Δ = 97
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{-(3)-\sqrt{97}}{2*-2}=\frac{-3-\sqrt{97}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{97}}{2*-2}=\frac{-3+\sqrt{97}}{-4} $
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