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10r^2-r-3=0
We add all the numbers together, and all the variables
10r^2-1r-3=0
a = 10; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·10·(-3)
Δ = 121
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{121}=11$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-11}{2*10}=\frac{-10}{20} =-1/2 $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*10}=\frac{12}{20} =3/5 $
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