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5x^2+12x=280
We move all terms to the left:
5x^2+12x-(280)=0
a = 5; b = 12; c = -280;
Δ = b2-4ac
Δ = 122-4·5·(-280)
Δ = 5744
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{5744}=\sqrt{16*359}=\sqrt{16}*\sqrt{359}=4\sqrt{359}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{359}}{2*5}=\frac{-12-4\sqrt{359}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{359}}{2*5}=\frac{-12+4\sqrt{359}}{10} $
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