10=2x^2+3x

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Solution for 10=2x^2+3x equation:



10=2x^2+3x
We move all terms to the left:
10-(2x^2+3x)=0
We get rid of parentheses
-2x^2-3x+10=0
a = -2; b = -3; c = +10;
Δ = b2-4ac
Δ = -32-4·(-2)·10
Δ = 89
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{89}}{2*-2}=\frac{3-\sqrt{89}}{-4} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{89}}{2*-2}=\frac{3+\sqrt{89}}{-4} $

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