2x^2+(14/5)x-228=0

Solution for 2x^2+(14/5)x-228=0 equation:

2x^2+(14/5)x-228=0


Domain of the equation: 5)x!=0

x!=0/1

x!=0

x∈R


We add all the numbers together, and all the variables

2x^2+(+14/5)x-228=0

We multiply parentheses

2x^2+14x^2-228=0

We add all the numbers together, and all the variables

16x^2-228=0

a = 16; b = 0; c = -228;
Δ = b2-4ac
Δ = 02-4·16·(-228)
Δ = 14592
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{14592}=\sqrt{256*57}=\sqrt{256}*\sqrt{57}=16\sqrt{57}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{57}}{2*16}=\frac{0-16\sqrt{57}}{32}
=-\frac{16\sqrt{57}}{32}
=-\frac{\sqrt{57}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{57}}{2*16}=\frac{0+16\sqrt{57}}{32}
=\frac{16\sqrt{57}}{32}
=\frac{\sqrt{57}}{2}
$