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10x^2-48x+32=0
a = 10; b = -48; c = +32;
Δ = b2-4ac
Δ = -482-4·10·32
Δ = 1024
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{1024}=32$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-32}{2*10}=\frac{16}{20} =4/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+32}{2*10}=\frac{80}{20} =4 $
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