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12x^2-23x+10=0
a = 12; b = -23; c = +10;
Δ = b2-4ac
Δ = -232-4·12·10
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-7}{2*12}=\frac{16}{24} =2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+7}{2*12}=\frac{30}{24} =1+1/4 $
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