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50=(2x+3)(x)
We move all terms to the left:
50-((2x+3)(x))=0
We calculate terms in parentheses: -((2x+3)x), so:We get rid of parentheses
(2x+3)x
We multiply parentheses
2x^2+3x
Back to the equation:
-(2x^2+3x)
-2x^2-3x+50=0
a = -2; b = -3; c = +50;
Δ = b2-4ac
Δ = -32-4·(-2)·50
Δ = 409
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{409}}{2*-2}=\frac{3-\sqrt{409}}{-4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{409}}{2*-2}=\frac{3+\sqrt{409}}{-4} $
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