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16x^2-148x+336=0
a = 16; b = -148; c = +336;
Δ = b2-4ac
Δ = -1482-4·16·336
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-148)-20}{2*16}=\frac{128}{32} =4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-148)+20}{2*16}=\frac{168}{32} =5+1/4 $
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