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6a^2-48a+42=0
a = 6; b = -48; c = +42;
Δ = b2-4ac
Δ = -482-4·6·42
Δ = 1296
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{1296}=36$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-36}{2*6}=\frac{12}{12} =1 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+36}{2*6}=\frac{84}{12} =7 $
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