3x^2 + 19x - 14 = 0, solve using the quadratic formula?

x = [ - 19 ± √ (361 + 168) ] / 6
x = [ - 19 ± √ (529) ] / 6
x = [ - 19 ± 23 ] / 6
x = 4 / 6 , x = - 42 / 6
x = 2/3 , x = - 7

use quadratic equation, with a=3, b=19, c= -14

For ax2 + bx + c = 0, the value of x is given by:

x = [ -b ± sqrt(b^2 - 4ac) ] / 2a

[-19 +/- sqrt (361 + 168)]/6

[-19 +/- sqrt (529)]/6

[-19 +/- 23]/6

-19 + 23 = 4
-19 - 23 = 42

4/6 = 2/3
42/6 = 7

answers: 2/3, 7

hi, 3x^2 + 19x – 14 =0 ?= b^2 – 4ac = 19^2 – 4(3)(-14) = 529; ??= ?529 = 23 x = (-b ± ??) /2a = (-19 ± 23) /2(3) x1 = (-19 + 23) /6 = 4/6 = 2/3 <<< answer! x2 = (-19 – 23) /6 = -7 <<< answer! hence, x1= 2/3 and x2 = -7. i desire that helps!!! :-)

3x^2 + 19x - 14 = 0
(3x - 2)(x + 7) = 0

3x - 2 = 0
3x = 2
x = 2/3

x + 7 = 0
x = -7

∴ x = 2/3 , -7

How do you not like the quadratic equation, it is beautiful,
anway you just use the coefficients like this
X1,2=(-19+ or -sqrt(19^2-4*3*-14))/(2*3) which sould give you two solutions.

3x^2+21x-2x -14=0

3x(x+7)- 2(x+7) =0

(3x-2) (x+7)= 0

x= -7, x= 2/3

x= 2/3, -7

3x^2 + 19x - 14 = 0

I don't like the quadratic formula. Here's the factored form:
(3x-2)(x+7) = 0
3x-2=0, 3x=2, x=2/3
x+7=0, x=-7