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2x^2+x-16=3x+8
We move all terms to the left:
2x^2+x-16-(3x+8)=0
We get rid of parentheses
2x^2+x-3x-8-16=0
We add all the numbers together, and all the variables
2x^2-2x-24=0
a = 2; b = -2; c = -24;
Δ = b2-4ac
Δ = -22-4·2·(-24)
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*2}=\frac{-12}{4} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*2}=\frac{16}{4} =4 $
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