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2.5=2x^2-3x
We move all terms to the left:
2.5-(2x^2-3x)=0
We get rid of parentheses
-2x^2+3x+2.5=0
a = -2; b = 3; c = +2.5;
Δ = b2-4ac
Δ = 32-4·(-2)·2.5
Δ = 29
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{29}}{2*-2}=\frac{-3-\sqrt{29}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{29}}{2*-2}=\frac{-3+\sqrt{29}}{-4} $
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