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5x^2+20x=6
We move all terms to the left:
5x^2+20x-(6)=0
a = 5; b = 20; c = -6;
Δ = b2-4ac
Δ = 202-4·5·(-6)
Δ = 520
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{130}}{2*5}=\frac{-20-2\sqrt{130}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{130}}{2*5}=\frac{-20+2\sqrt{130}}{10} $
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