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-3x^2+20x+23=0
a = -3; b = 20; c = +23;
Δ = b2-4ac
Δ = 202-4·(-3)·23
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-26}{2*-3}=\frac{-46}{-6} =7+2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+26}{2*-3}=\frac{6}{-6} =-1 $
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