How do you solve $x^{2} - 4 x - 3 = 0$ $x^2 - 4x -3 = 0$ ?

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Answer 1 · 2015-08-09T20:55:28

Use the quadratic formula to find $x = 2 \pm \sqrt{7}$ $x = 2+-sqrt(7)$

$x^{2} - 4 x - 3$ $x^2-4x-3$ is of the form $a x^{2} + b x + c$ $ax^2+bx+c$ , with $a = 1$ $a=1$ , $b = - 4$ $b=-4$ and $c = - 3$ $c=-3$ .

This has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = (-4)^2-(4xx1xx-3) = 16+12 = 28$

$= 2^2*7$

This is positive but not a perfect square, so $x^2-4x-3 = 0$ has two distinct irrational roots.

Use the quadratic formula to give the roots:

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

$= (4+-2sqrt(7))/2 = 2+-sqrt(7)$

How do you solve x^2 - 4x -3 = 0x2−4x−3=0x^2 - 4x -3 = 0?