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(2x-19)(x-6)=14
We move all terms to the left:
(2x-19)(x-6)-(14)=0
We multiply parentheses ..
(+2x^2-12x-19x+114)-14=0
We get rid of parentheses
2x^2-12x-19x+114-14=0
We add all the numbers together, and all the variables
2x^2-31x+100=0
a = 2; b = -31; c = +100;
Δ = b2-4ac
Δ = -312-4·2·100
Δ = 161
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{-(-31)-\sqrt{161}}{2*2}=\frac{31-\sqrt{161}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{161}}{2*2}=\frac{31+\sqrt{161}}{4} $
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