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12x^2+21x-106=0
a = 12; b = 21; c = -106;
Δ = b2-4ac
Δ = 212-4·12·(-106)
Δ = 5529
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{5529}}{2*12}=\frac{-21-\sqrt{5529}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{5529}}{2*12}=\frac{-21+\sqrt{5529}}{24} $
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