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3x^2-4x=x+50
We move all terms to the left:
3x^2-4x-(x+50)=0
We get rid of parentheses
3x^2-4x-x-50=0
We add all the numbers together, and all the variables
3x^2-5x-50=0
a = 3; b = -5; c = -50;
Δ = b2-4ac
Δ = -52-4·3·(-50)
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-25}{2*3}=\frac{-20}{6} =-3+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+25}{2*3}=\frac{30}{6} =5 $
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