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16r^2-38r+21=0
a = 16; b = -38; c = +21;
Δ = b2-4ac
Δ = -382-4·16·21
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-10}{2*16}=\frac{28}{32} =7/8 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+10}{2*16}=\frac{48}{32} =1+1/2 $
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