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4x(5x-3)=21
We move all terms to the left:
4x(5x-3)-(21)=0
We multiply parentheses
20x^2-12x-21=0
a = 20; b = -12; c = -21;
Δ = b2-4ac
Δ = -122-4·20·(-21)
Δ = 1824
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1824}=\sqrt{16*114}=\sqrt{16}*\sqrt{114}=4\sqrt{114}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{114}}{2*20}=\frac{12-4\sqrt{114}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{114}}{2*20}=\frac{12+4\sqrt{114}}{40} $
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