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3x^2-25x+22=0
a = 3; b = -25; c = +22;
Δ = b2-4ac
Δ = -252-4·3·22
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-19}{2*3}=\frac{6}{6} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+19}{2*3}=\frac{44}{6} =7+1/3 $
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