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3x^2+8x-256=0
a = 3; b = 8; c = -256;
Δ = b2-4ac
Δ = 82-4·3·(-256)
Δ = 3136
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{3136}=56$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-56}{2*3}=\frac{-64}{6} =-10+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+56}{2*3}=\frac{48}{6} =8 $
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