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-6x(-2x+3)=42
We move all terms to the left:
-6x(-2x+3)-(42)=0
We multiply parentheses
12x^2-18x-42=0
a = 12; b = -18; c = -42;
Δ = b2-4ac
Δ = -182-4·12·(-42)
Δ = 2340
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2340}=\sqrt{36*65}=\sqrt{36}*\sqrt{65}=6\sqrt{65}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{65}}{2*12}=\frac{18-6\sqrt{65}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{65}}{2*12}=\frac{18+6\sqrt{65}}{24} $
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