2x^2+35x+48=0

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Solution for 2x^2+35x+48=0 equation:



2x^2+35x+48=0
a = 2; b = 35; c = +48;
Δ = b2-4ac
Δ = 352-4·2·48
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-29}{2*2}=\frac{-64}{4} =-16 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+29}{2*2}=\frac{-6}{4} =-1+1/2 $

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