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3x^2+4x-57=0
a = 3; b = 4; c = -57;
Δ = b2-4ac
Δ = 42-4·3·(-57)
Δ = 700
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-10\sqrt{7}}{2*3}=\frac{-4-10\sqrt{7}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+10\sqrt{7}}{2*3}=\frac{-4+10\sqrt{7}}{6} $
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