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12x^2-26x-7=3x+7
We move all terms to the left:
12x^2-26x-7-(3x+7)=0
We get rid of parentheses
12x^2-26x-3x-7-7=0
We add all the numbers together, and all the variables
12x^2-29x-14=0
a = 12; b = -29; c = -14;
Δ = b2-4ac
Δ = -292-4·12·(-14)
Δ = 1513
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{-(-29)-\sqrt{1513}}{2*12}=\frac{29-\sqrt{1513}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{1513}}{2*12}=\frac{29+\sqrt{1513}}{24} $
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