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17=(2x-4)(3x-5)
We move all terms to the left:
17-((2x-4)(3x-5))=0
We multiply parentheses ..
-((+6x^2-10x-12x+20))+17=0
We calculate terms in parentheses: -((+6x^2-10x-12x+20)), so:We get rid of parentheses
(+6x^2-10x-12x+20)
We get rid of parentheses
6x^2-10x-12x+20
We add all the numbers together, and all the variables
6x^2-22x+20
Back to the equation:
-(6x^2-22x+20)
-6x^2+22x-20+17=0
We add all the numbers together, and all the variables
-6x^2+22x-3=0
a = -6; b = 22; c = -3;
Δ = b2-4ac
Δ = 222-4·(-6)·(-3)
Δ = 412
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{103}}{2*-6}=\frac{-22-2\sqrt{103}}{-12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{103}}{2*-6}=\frac{-22+2\sqrt{103}}{-12} $
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