Can you Help Please?

Question

I'm stuck on this math problem can u help I'm bad with quadratic formula

use the quadratic formula to solve this equation

x^2 +5x-14=0

Answer 1

ax² + bx + c = 0
1x² + 5x - 14 = 0 <-- Your equation.

Quadratic formula:
x = [-b ± √(b² - 4ac)] / 2a
x = [-5 ± √(5² - 4(1)(-14))] / 2(1)
x = [-5 ± √(25 + 56)] / 2
x = [-5 ± √(81)] / 2
x = [-5 ± 9] / 2

So, 2 anwers:
x = (-5 + 9)/2, and x = (-5 - 9)/2
x = (4)/2, and x = (-14)/2
x = 2, and x = -7

Answer 2

It is a mere substitution. Go back and master this

aX^2 + bx + c =0


x = [-b +/-(b^2 - 4ac)^0.5]/2a

in your case

x = [-5 =/- (14^2 - 4*1*(-14))^0.5]/2*1

This is very straight forward; you have not mastered the material and it has to be mastered to proceed. In math you just cannot allow yourself to miss a step. You will use the quadratic formula many times in the future.

Answer 3

x^2 + 5x - 14 = 0

x = {-b +/- (b^2 - 4ac)^1/2 }/2a
where, a = 1, b = 5, c = -14.

x = {-5 +/- (25 - (4*1*-14)^1/2 }/2
x = {-5 +/- (81)^1/2 }/2
x = (-5 +/- 9)/2

x = (-5 + 9)/2 = 2
x = (-5 - 9)/2 = -7

x = 2, x = -7

Answer 4

2ax = -b +/- (b^2 - 4ac)^1/2

a = 1
b = 5
c = -14

x = [-5 +/- (25 - 4(1)(-14))^1/2] /2
x = [-5 +/- (81)^(1/2)]/2
x = [-5 +/ 9] /2
x = 2, -7

Answer 5

-5 + or - (sqr rt( 25 - (-56) ) ) / 2

= -5 + or - (sqr rt(81) ) / 2

= -5 + or - (9/2)

= -5 + or - 4.5

= -1.5 or -9.5