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3x(4x+5)=33
We move all terms to the left:
3x(4x+5)-(33)=0
We multiply parentheses
12x^2+15x-33=0
a = 12; b = 15; c = -33;
Δ = b2-4ac
Δ = 152-4·12·(-33)
Δ = 1809
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1809}=\sqrt{9*201}=\sqrt{9}*\sqrt{201}=3\sqrt{201}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{201}}{2*12}=\frac{-15-3\sqrt{201}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{201}}{2*12}=\frac{-15+3\sqrt{201}}{24} $
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