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4(3x+5)*x=11
We move all terms to the left:
4(3x+5)*x-(11)=0
We multiply parentheses
12x^2+20x-11=0
a = 12; b = 20; c = -11;
Δ = b2-4ac
Δ = 202-4·12·(-11)
Δ = 928
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{928}=\sqrt{16*58}=\sqrt{16}*\sqrt{58}=4\sqrt{58}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{58}}{2*12}=\frac{-20-4\sqrt{58}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{58}}{2*12}=\frac{-20+4\sqrt{58}}{24} $
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