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2x^2+5x-10=x^2+14
We move all terms to the left:
2x^2+5x-10-(x^2+14)=0
We get rid of parentheses
2x^2-x^2+5x-14-10=0
We add all the numbers together, and all the variables
x^2+5x-24=0
a = 1; b = 5; c = -24;
Δ = b2-4ac
Δ = 52-4·1·(-24)
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*1}=\frac{-16}{2} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*1}=\frac{6}{2} =3 $
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