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(3x-4)(2x-10)=10
We move all terms to the left:
(3x-4)(2x-10)-(10)=0
We multiply parentheses ..
(+6x^2-30x-8x+40)-10=0
We get rid of parentheses
6x^2-30x-8x+40-10=0
We add all the numbers together, and all the variables
6x^2-38x+30=0
a = 6; b = -38; c = +30;
Δ = b2-4ac
Δ = -382-4·6·30
Δ = 724
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{724}=\sqrt{4*181}=\sqrt{4}*\sqrt{181}=2\sqrt{181}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{181}}{2*6}=\frac{38-2\sqrt{181}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{181}}{2*6}=\frac{38+2\sqrt{181}}{12} $
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