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2x^2+4x-7+5x-11=180
We move all terms to the left:
2x^2+4x-7+5x-11-(180)=0
We add all the numbers together, and all the variables
2x^2+9x-198=0
a = 2; b = 9; c = -198;
Δ = b2-4ac
Δ = 92-4·2·(-198)
Δ = 1665
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1665}=\sqrt{9*185}=\sqrt{9}*\sqrt{185}=3\sqrt{185}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{185}}{2*2}=\frac{-9-3\sqrt{185}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{185}}{2*2}=\frac{-9+3\sqrt{185}}{4} $
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