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2x^2+5x-3=x+27
We move all terms to the left:
2x^2+5x-3-(x+27)=0
We get rid of parentheses
2x^2+5x-x-27-3=0
We add all the numbers together, and all the variables
2x^2+4x-30=0
a = 2; b = 4; c = -30;
Δ = b2-4ac
Δ = 42-4·2·(-30)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*2}=\frac{-20}{4} =-5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*2}=\frac{12}{4} =3 $
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