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(2-3x)(x+11)=(3x-2)(2-5x)
We move all terms to the left:
(2-3x)(x+11)-((3x-2)(2-5x))=0
We add all the numbers together, and all the variables
(-3x+2)(x+11)-((3x-2)(-5x+2))=0
We multiply parentheses ..
(-3x^2-33x+2x+22)-((3x-2)(-5x+2))=0
We calculate terms in parentheses: -((3x-2)(-5x+2)), so:We get rid of parentheses
(3x-2)(-5x+2)
We multiply parentheses ..
(-15x^2+6x+10x-4)
We get rid of parentheses
-15x^2+6x+10x-4
We add all the numbers together, and all the variables
-15x^2+16x-4
Back to the equation:
-(-15x^2+16x-4)
-3x^2+15x^2-33x+2x-16x+22+4=0
We add all the numbers together, and all the variables
12x^2-47x+26=0
a = 12; b = -47; c = +26;
Δ = b2-4ac
Δ = -472-4·12·26
Δ = 961
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{961}=31$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-47)-31}{2*12}=\frac{16}{24} =2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-47)+31}{2*12}=\frac{78}{24} =3+1/4 $
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