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5a^2+22a-48=0
a = 5; b = 22; c = -48;
Δ = b2-4ac
Δ = 222-4·5·(-48)
Δ = 1444
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{1444}=38$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-38}{2*5}=\frac{-60}{10} =-6 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+38}{2*5}=\frac{16}{10} =1+3/5 $
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