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5x+2x^2=4x^2-9x
We move all terms to the left:
5x+2x^2-(4x^2-9x)=0
We get rid of parentheses
2x^2-4x^2+5x+9x=0
We add all the numbers together, and all the variables
-2x^2+14x=0
a = -2; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·(-2)·0
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*-2}=\frac{-28}{-4} =+7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*-2}=\frac{0}{-4} =0 $
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