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48x^2+22x-5=0
a = 48; b = 22; c = -5;
Δ = b2-4ac
Δ = 222-4·48·(-5)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1444}=38$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-38}{2*48}=\frac{-60}{96} =-5/8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+38}{2*48}=\frac{16}{96} =1/6 $
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