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16x^2+52x+42=0
a = 16; b = 52; c = +42;
Δ = b2-4ac
Δ = 522-4·16·42
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(52)-4}{2*16}=\frac{-56}{32} =-1+3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(52)+4}{2*16}=\frac{-48}{32} =-1+1/2 $
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