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12x^2-29x+10=0
a = 12; b = -29; c = +10;
Δ = b2-4ac
Δ = -292-4·12·10
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-19}{2*12}=\frac{10}{24} =5/12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+19}{2*12}=\frac{48}{24} =2 $
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