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4x(3x+11)=60
We move all terms to the left:
4x(3x+11)-(60)=0
We multiply parentheses
12x^2+44x-60=0
a = 12; b = 44; c = -60;
Δ = b2-4ac
Δ = 442-4·12·(-60)
Δ = 4816
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{4816}=\sqrt{16*301}=\sqrt{16}*\sqrt{301}=4\sqrt{301}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(44)-4\sqrt{301}}{2*12}=\frac{-44-4\sqrt{301}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(44)+4\sqrt{301}}{2*12}=\frac{-44+4\sqrt{301}}{24} $
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