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12x^2+17x=6
We move all terms to the left:
12x^2+17x-(6)=0
a = 12; b = 17; c = -6;
Δ = b2-4ac
Δ = 172-4·12·(-6)
Δ = 577
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{-(17)-\sqrt{577}}{2*12}=\frac{-17-\sqrt{577}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{577}}{2*12}=\frac{-17+\sqrt{577}}{24} $
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