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10a^2-23a+12=0
a = 10; b = -23; c = +12;
Δ = b2-4ac
Δ = -232-4·10·12
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{49}=7$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-7}{2*10}=\frac{16}{20} =4/5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+7}{2*10}=\frac{30}{20} =1+1/2 $
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