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3x^2+10x-48=0
a = 3; b = 10; c = -48;
Δ = b2-4ac
Δ = 102-4·3·(-48)
Δ = 676
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{676}=26$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-26}{2*3}=\frac{-36}{6} =-6 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+26}{2*3}=\frac{16}{6} =2+2/3 $
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