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14x^2+3-23x=0
a = 14; b = -23; c = +3;
Δ = b2-4ac
Δ = -232-4·14·3
Δ = 361
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{361}=19$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-19}{2*14}=\frac{4}{28} =1/7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+19}{2*14}=\frac{42}{28} =1+1/2 $
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