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2x^2+x+12x-5=22
We move all terms to the left:
2x^2+x+12x-5-(22)=0
We add all the numbers together, and all the variables
2x^2+13x-27=0
a = 2; b = 13; c = -27;
Δ = b2-4ac
Δ = 132-4·2·(-27)
Δ = 385
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{-(13)-\sqrt{385}}{2*2}=\frac{-13-\sqrt{385}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{385}}{2*2}=\frac{-13+\sqrt{385}}{4} $
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