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3x^2+28x+64=0
a = 3; b = 28; c = +64;
Δ = b2-4ac
Δ = 282-4·3·64
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4}{2*3}=\frac{-32}{6} =-5+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4}{2*3}=\frac{-24}{6} =-4 $
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