Question

Solve using the quadratic formula 2x^2+8x-5=0

Answer 1

Step-by-step explanation:

Hey there!!

Given,

${x}^{2} + 8x - 5 = 0$

Comparing it with ax^2+bx+c =0 we get,

a= 1, b= 8, c= -5.

Using formula,

$x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a}$

Keep all values,

$x = \frac{ - 8 + - \sqrt{ {8}^{2} - 4 \times 1 \times ( - 5)} }{2 \times 1}$

$x = \frac{ - 8 + - \sqrt{64 + 20} }{2}$

$x = \frac{ - 8 + - \sqrt{84} }{2}$

$x = \frac{ - 8 + - 2 \sqrt{21} }{2}$

Taking negative,

$x = \frac{ - 8 - 2 \sqrt{21} }{2}$

$x = \frac{ - 2(4 + \sqrt{21}) }{2}$

$x = 4 + \sqrt{21}$

Similarly, taking positive,

$x = \frac{ - 8 + 2 \sqrt{21} }{2}$

$x = \frac{ - 2(4 - \sqrt{21} ) }{2}$

$x = 4 - \sqrt{21}$

Therefore, x= (4 + root 21, 4 - root 21).

Hope it helps.....