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2x^2+4x-24=5x+12
We move all terms to the left:
2x^2+4x-24-(5x+12)=0
We get rid of parentheses
2x^2+4x-5x-12-24=0
We add all the numbers together, and all the variables
2x^2-1x-36=0
a = 2; b = -1; c = -36;
Δ = b2-4ac
Δ = -12-4·2·(-36)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-17}{2*2}=\frac{-16}{4} =-4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+17}{2*2}=\frac{18}{4} =4+1/2 $
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