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10=2x^2+9x
We move all terms to the left:
10-(2x^2+9x)=0
We get rid of parentheses
-2x^2-9x+10=0
a = -2; b = -9; c = +10;
Δ = b2-4ac
Δ = -92-4·(-2)·10
Δ = 161
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{-(-9)-\sqrt{161}}{2*-2}=\frac{9-\sqrt{161}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{161}}{2*-2}=\frac{9+\sqrt{161}}{-4} $
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