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(3x-1)(x-3)-(x+4)(x-2)=3(x+4)
We move all terms to the left:
(3x-1)(x-3)-(x+4)(x-2)-(3(x+4))=0
We multiply parentheses ..
(+3x^2-9x-1x+3)-(x+4)(x-2)-(3(x+4))=0
We calculate terms in parentheses: -(3(x+4)), so:We get rid of parentheses
3(x+4)
We multiply parentheses
3x+12
Back to the equation:
-(3x+12)
3x^2-9x-1x-(x+4)(x-2)-3x+3-12=0
We multiply parentheses ..
3x^2-(+x^2-2x+4x-8)-9x-1x-3x+3-12=0
We add all the numbers together, and all the variables
3x^2-(+x^2-2x+4x-8)-13x-9=0
We get rid of parentheses
3x^2-x^2+2x-4x-13x+8-9=0
We add all the numbers together, and all the variables
2x^2-15x-1=0
a = 2; b = -15; c = -1;
Δ = b2-4ac
Δ = -152-4·2·(-1)
Δ = 233
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{-(-15)-\sqrt{233}}{2*2}=\frac{15-\sqrt{233}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{233}}{2*2}=\frac{15+\sqrt{233}}{4} $
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