How do you solve #6x^2-8x-3=0# using the quadratic formula?

Question

Eli Assouline · Accepted Answer

undefined To solve the quadratic equation $6x^2 - 8x - 3 = 0$ using the quadratic formula:

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

where $a = 6$, $b = -8$, and $c = -3$:

$$x = \frac{{-(-8) \pm \sqrt{{(-8)^2 - 4 \cdot 6 \cdot (-3)}}}}{{2 \cdot 6}}$$

$$x = \frac{{8 \pm \sqrt{{64 + 72}}}}{{12}}$$

$$x = \frac{{8 \pm \sqrt{{136}}}}{{12}}$$

$$x = \frac{{8 \pm 2\sqrt{{34}}}}{{12}}$$

$$x = \frac{{4 \pm \sqrt{{34}}}}{{6}}$$

So, the solutions for the equation $6x^2 - 8x - 3 = 0$ are $x = \frac{{4 + \sqrt{{34}}}}{{6}}$ and $x = \frac{{4 - \sqrt{{34}}}}{{6}}$.

Nathan Axel · Answer

#x=(4+-sqrt(34))/6# #1#. Since the given equation is already in standard form, identify the #color(blue)a,color(darkorange)b,# and #color(violet)c# values. Then plug the values into the quadratic formula to solve for the roots. #color(blue)6x^2# #color(darkorange)(-8)x# #color(violet)(-3)=0# #color(blue)(a=6)color(white)(XXXXX)color(darkorange)(b=-8)color(white)(XXXXX)color(violet)(c=-3)# #x=(-b+-sqrt(b^2-4ac))/(2a)# #x=(-(color(darkorange)(-8))+-sqrt((color(darkorange)(-8))^2-4(color(blue)6)(color(violet)(-3))))/(2(color(blue)6))# #x=(8+-sqrt(64+72))/12# #x=(8+-sqrt(136))/12# #x=(8+-2sqrt(34))/12# #2#. Factor out #2# from the numerator and denominator. #x=(2(4+-sqrt(34)))/(2(6))# #x=(color(red)cancelcolor(black)2(4+-sqrt(34)))/(color(red)cancelcolor(black)2(6))# #color(green)(|bar(ul(color(white)(a/a)x=(4+-sqrt(34))/6color(white)(a/a)|)))#

Ethan Adkins · Answer

#x_(1,2) = (-color(blue)8+- sqrt(color(blue)(64)-72))/(12)= -2/3 +- isqrt(2)/6 = 1/3(-2+-sqrt(2)/2)#
Note this means:
#x_1 = 1/3 (-2+sqrt(2)/2)# and #x_1 = 1/3 (-2-sqrt(2)/2)# I suggest you commit to memory the "Quadratic Formula" it probably one of the formula that you absolutely positively must know by heart: So here is your Quadratic Formula: Given a 2nd Order Polynomial,#P_2#: #P_2 = color(red)ax^2 + color(blue)bx + color(green)c# the Roots or solutions to the equation #x_1# and #x_2# are given by the Quadratic Formula: #x_(1,2) = (-color(blue)b+- sqrt(color(blue)(b^2)-4color(red)acolor(green)c))/(2color(red)a)# Now for your equation: #color(red)6x^2-color(blue)8x-color(green)3=0# #x_(1,2) = (-color(blue)8+- sqrt(color(blue)(8^2)-4*color(red)6*color(green)3))/(2*color(red)6)# #x_(1,2) = (-color(blue)8+- sqrt(color(blue)(64)-72))/(12)= -2/3 +- isqrt(2)/6 = 1/3(-2+-sqrt(2)/2)#