How do you work out these question using the quadratic formula?

x^2 + 4x + 3 = 0 <------------> ax^2 + bx + constant = 0

Note: a = 1, b = 4, c = 3

Quadratic formula: x = -(b) ± [(b)^2 - 4(a)(c)]^(1/2) .... all over 2(a)

x = -(4) ± [(4)^2 - 4(1)(3]^(1/2) ... all over 2(1)

x = [-4 ± √(4)] / 2

x = [-4 ± 2] / 2

You get one value of x when you add the 2:

x = -1

You get another value of x when you subtract the 2:

x = -3

Check work: x^2 + 4x + 3 = 0

(-1)^2 + 4(-1) + 3 ?=? 0 <--------> 1 - 4 + 3 ?=? 0 <--------> 0 = 0 Correct!

(-3)^2 + 4(-3) + 3 ?=? 0 <--------> 9 - 12 + 3 ?=? 0 <-------> 0 = 0 Correct!
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by using the formula,
x = [-b +- sqrt(b^2 - 4ac) ] / 2a
x = -4 ± √[4² - 4(1)(3)] /2(1)
x = -4 ± 2 / 2
x = -2 ± 1
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Well,
in this special case, you also can see that in :

x^2 + 4x + 3 =0 ==> -1 is an obvious solution

therefore, the other is -c/a = -3/1 = -3

hope it' ll help !!
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By using the quadratic formula.

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

Write down a, b and c and plug them into the formula. I'm not sure what other information you need to know.

a is the coefficient of x^2.
b is the coefficient of x
c is the constant term.

Look at your equation. Write them down. Plug them in.
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x = -4 ± √[4² - 4(1)(3)] ÷ 2(1)
x = -4 ± 2 ÷ 2
x = -2 ± 1
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yes
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x = [-4 +/- sqrt (4² - 3 x 4)]/2 = 2 +/- 1 = 1 or 3.