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2x^2+8x+18=42
We move all terms to the left:
2x^2+8x+18-(42)=0
We add all the numbers together, and all the variables
2x^2+8x-24=0
a = 2; b = 8; c = -24;
Δ = b2-4ac
Δ = 82-4·2·(-24)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-16}{2*2}=\frac{-24}{4} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+16}{2*2}=\frac{8}{4} =2 $
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