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4x(1x-3)=20
We move all terms to the left:
4x(1x-3)-(20)=0
We add all the numbers together, and all the variables
4x(x-3)-20=0
We multiply parentheses
4x^2-12x-20=0
a = 4; b = -12; c = -20;
Δ = b2-4ac
Δ = -122-4·4·(-20)
Δ = 464
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{464}=\sqrt{16*29}=\sqrt{16}*\sqrt{29}=4\sqrt{29}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{29}}{2*4}=\frac{12-4\sqrt{29}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{29}}{2*4}=\frac{12+4\sqrt{29}}{8} $
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