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15=2x^2-x
We move all terms to the left:
15-(2x^2-x)=0
We get rid of parentheses
-2x^2+x+15=0
a = -2; b = 1; c = +15;
Δ = b2-4ac
Δ = 12-4·(-2)·15
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*-2}=\frac{-12}{-4} =+3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*-2}=\frac{10}{-4} =-2+1/2 $
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