How do you solve #8x^2-38x=4# using the quadratic formula?

Question

Nathan Axel · Accepted Answer

#x=-0.1025#
#x=4.85#

Given - #8x^2-38x-4=0# #x=(-b+-sqrt(b^2-4ac))/(2a)# #a = 8# #b=-38# #c=-4# #x=(-(38)+-sqrt(-38^2-(4xx8xx-4)))/(2xx8# #x=(38+-sqrt(1444-(-128)))/16# #x=(38+-sqrt(1572))/16# #x=(38+-39.64)/16# #x=(38-39.64)/16=(-1.64)/16=-0.1025# #x=(38+39.64)/16=(77.64)/16=4.85# #x=-0.1025# #x=4.85#

Savannah Anderson · Answer

undefined To solve the quadratic equation $8x^2 - 38x = 4$ using the quadratic formula, where $ax^2 + bx + c = 0$, the quadratic formula is:

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

For the given equation, $a = 8$, $b = -38$, and $c = -4$.

Substitute these values into the quadratic formula:

$$x = \frac{{-(-38) \pm \sqrt{{(-38)^2 - 4(8)(-4)}}}}{{2(8)}}$$

$$x = \frac{{38 \pm \sqrt{{1444 + 128}}}}{{16}}$$

$$x = \frac{{38 \pm \sqrt{{1572}}}}{{16}}$$

$$x = \frac{{38 \pm 2\sqrt{{393}}}}{{16}}$$

Therefore, the solutions to the equation $8x^2 - 38x = 4$ are:

$$x = \frac{{38 + 2\sqrt{{393}}}}{{16}}$$

$$x = \frac{{38 - 2\sqrt{{393}}}}{{16}}$$