4x(3x+40)=20

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Solution for 4x(3x+40)=20 equation:



4x(3x+40)=20
We move all terms to the left:
4x(3x+40)-(20)=0
We multiply parentheses
12x^2+160x-20=0
a = 12; b = 160; c = -20;
Δ = b2-4ac
Δ = 1602-4·12·(-20)
Δ = 26560
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:
$\sqrt{\Delta}=\sqrt{26560}=\sqrt{64*415}=\sqrt{64}*\sqrt{415}=8\sqrt{415}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-8\sqrt{415}}{2*12}=\frac{-160-8\sqrt{415}}{24} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+8\sqrt{415}}{2*12}=\frac{-160+8\sqrt{415}}{24} $

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