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5a^2+a-48=0
a = 5; b = 1; c = -48;
Δ = b2-4ac
Δ = 12-4·5·(-48)
Δ = 961
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{961}=31$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*5}=\frac{-32}{10} =-3+1/5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*5}=\frac{30}{10} =3 $
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