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10=x(3x-13)
We move all terms to the left:
10-(x(3x-13))=0
We calculate terms in parentheses: -(x(3x-13)), so:We get rid of parentheses
x(3x-13)
We multiply parentheses
3x^2-13x
Back to the equation:
-(3x^2-13x)
-3x^2+13x+10=0
a = -3; b = 13; c = +10;
Δ = b2-4ac
Δ = 132-4·(-3)·10
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*-3}=\frac{-30}{-6} =+5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*-3}=\frac{4}{-6} =-2/3 $
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