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21x^2+60x-120=0
a = 21; b = 60; c = -120;
Δ = b2-4ac
Δ = 602-4·21·(-120)
Δ = 13680
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{13680}=\sqrt{144*95}=\sqrt{144}*\sqrt{95}=12\sqrt{95}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-12\sqrt{95}}{2*21}=\frac{-60-12\sqrt{95}}{42} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+12\sqrt{95}}{2*21}=\frac{-60+12\sqrt{95}}{42} $
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