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4x(5x+3)=19
We move all terms to the left:
4x(5x+3)-(19)=0
We multiply parentheses
20x^2+12x-19=0
a = 20; b = 12; c = -19;
Δ = b2-4ac
Δ = 122-4·20·(-19)
Δ = 1664
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{26}}{2*20}=\frac{-12-8\sqrt{26}}{40} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{26}}{2*20}=\frac{-12+8\sqrt{26}}{40} $
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