How do you solve the quadratic $x^{2} + 3 x - 28 = 0$ $x^2+3x-28=0$ using any method?

Question

8651 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-31T12:48:07

$x = 4, - 7$ $x= 4, -7$

$x^{2} + 3 x - 28 = 0$ $x^2 +3 x -28 = 0$

$x^{2} + 7 x - 4 x - 28 = 0$ $x^2 +7 x - 4 x -28= 0$

$x (x + 7) - 4 (x + 7) = 0$ $x ( x+7 ) -4 (x + 7 ) = 0$

$(x + 7) (x - 4) = 0$ $( x+7) ( x-4 ) = 0$

Either $(x + 7) = 0, or (x - 4) = 0$ $( x+7) = 0, or ( x-4) = 0$

If $x + 7 = 0$ $x+7 = 0$

$x = - 7$ $x = -7$

If $x - 7 = 0$ $x-7 = 0$

$x = 4$ $x= 4$

$x = 4, - 7$ $x = 4 , -7$

How do you solve the quadratic x2+3x−28=0x^2+3x-28=0 using any method?