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2x^2-18+3x^2+11x=180
We move all terms to the left:
2x^2-18+3x^2+11x-(180)=0
We add all the numbers together, and all the variables
5x^2+11x-198=0
a = 5; b = 11; c = -198;
Δ = b2-4ac
Δ = 112-4·5·(-198)
Δ = 4081
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{-(11)-\sqrt{4081}}{2*5}=\frac{-11-\sqrt{4081}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{4081}}{2*5}=\frac{-11+\sqrt{4081}}{10} $
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