How do you solve $x^2+4x+3=0$ using the quadratic formula?

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Answer 1 · 2017-07-22T12:35:39

$-1" or "-3$

For the quadratic equation

$ax^2+bx+c=0$

the formula for the roots is

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

$(1)" "$ Identify the values for $a,b,&" "c$

$x^2+4x+3=0$

cmp

$ax^2+bx+c=0$

$color(red)(a=1)$

$color(blue)(b=4)$

$color(green)(c=3)$

$(2)" "$ Substitute these numbers into eh formula

$x=(color(blue)(-4)+-sqrt(color(blue)(4^2)-(4xxcolor(red)(1)xxcolor(green)(3))))/(2xxcolor(red)(1))$

$(3)" "$ Carefully proceed and do the calculations

$x=(-4+-sqrt(16-12))/2$

$x=(-4+-sqrt4)/2$

$x=(-4+-2)/2$

now calculate the two separate solutions

$x_1=(-4+2)/2=-2/2=-1$

$x_2=(-4-2)/2=-6/2=-3$

How do you solve x^2+4x+3=0x^2+4x+3=0 using the quadratic formula?