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-9=2x^2+3x-10
We move all terms to the left:
-9-(2x^2+3x-10)=0
We get rid of parentheses
-2x^2-3x+10-9=0
We add all the numbers together, and all the variables
-2x^2-3x+1=0
a = -2; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·(-2)·1
Δ = 17
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{17}}{2*-2}=\frac{3-\sqrt{17}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{17}}{2*-2}=\frac{3+\sqrt{17}}{-4} $
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