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2x(3x-4)=10+3x
We move all terms to the left:
2x(3x-4)-(10+3x)=0
We add all the numbers together, and all the variables
2x(3x-4)-(3x+10)=0
We multiply parentheses
6x^2-8x-(3x+10)=0
We get rid of parentheses
6x^2-8x-3x-10=0
We add all the numbers together, and all the variables
6x^2-11x-10=0
a = 6; b = -11; c = -10;
Δ = b2-4ac
Δ = -112-4·6·(-10)
Δ = 361
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{361}=19$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-19}{2*6}=\frac{-8}{12} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*6}=\frac{30}{12} =2+1/2 $
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