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(x^2-3x-1)-(12x+6)=0
We get rid of parentheses
x^2-3x-12x-1-6=0
We add all the numbers together, and all the variables
x^2-15x-7=0
a = 1; b = -15; c = -7;
Δ = b2-4ac
Δ = -152-4·1·(-7)
Δ = 253
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{253}}{2*1}=\frac{15-\sqrt{253}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{253}}{2*1}=\frac{15+\sqrt{253}}{2} $
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