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4x^2+5x=9+2x^2
We move all terms to the left:
4x^2+5x-(9+2x^2)=0
We get rid of parentheses
4x^2-2x^2+5x-9=0
We add all the numbers together, and all the variables
2x^2+5x-9=0
a = 2; b = 5; c = -9;
Δ = b2-4ac
Δ = 52-4·2·(-9)
Δ = 97
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{-(5)-\sqrt{97}}{2*2}=\frac{-5-\sqrt{97}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{97}}{2*2}=\frac{-5+\sqrt{97}}{4} $
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