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54x^2+168x+120=0
a = 54; b = 168; c = +120;
Δ = b2-4ac
Δ = 1682-4·54·120
Δ = 2304
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2304}=48$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(168)-48}{2*54}=\frac{-216}{108} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(168)+48}{2*54}=\frac{-120}{108} =-1+1/9 $
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