# How do you solve the quadratic using the quadratic formula given 2a^2-6a-3=0 over the set of complex numbers?

May 13, 2017

The solution does not require complex numbers.

#### Explanation:

You simply put the coefficients of the equation into the formula and solve it algebraically.
For ${a}^{x} 2 + b \cdot x + c = 0$, the values of x which are the solutions of the equation are given by:

x = ​(−b±√[​b​^2​​−4ac])/(2a)

In this case, a = 2, b = -6 and c = -3

x = ​−(-6) ± sqrt[((-6)^​2)​ ​− (4*2*-3))/(2*2)

x = ​6 ± sqrt[(36 + 24])/4 ; x = ​(6 ± sqrt[60])/4
x = ​(6 ± 7.75)/4

$x = - 0.4375 , \mathmr{and} x = 3.4375$

CHECK : 2(-0.4375^2) – 6(-0.4375) -3 = 0 ; 2*(0.19) + 2.62 – 3 = 0 ;
0.38 + 2.62 – 3 = 0 ; 0 = 0 correct.