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2x^2+22x+48=0
a = 2; b = 22; c = +48;
Δ = b2-4ac
Δ = 222-4·2·48
Δ = 100
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{100}=10$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-10}{2*2}=\frac{-32}{4} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+10}{2*2}=\frac{-12}{4} =-3 $
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