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(5x)^2+16x-192=0
a = 5; b = 16; c = -192;
Δ = b2-4ac
Δ = 162-4·5·(-192)
Δ = 4096
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{4096}=64$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-64}{2*5}=\frac{-80}{10} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+64}{2*5}=\frac{48}{10} =4+4/5 $
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