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4(3x+5)x=5
We move all terms to the left:
4(3x+5)x-(5)=0
We multiply parentheses
12x^2+20x-5=0
a = 12; b = 20; c = -5;
Δ = b2-4ac
Δ = 202-4·12·(-5)
Δ = 640
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8\sqrt{10}}{2*12}=\frac{-20-8\sqrt{10}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8\sqrt{10}}{2*12}=\frac{-20+8\sqrt{10}}{24} $
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