How do you solve $x^2 + 2x-7=0$ using the quadratic formula?

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Answer 1 · 2016-02-14T09:41:52

$color(green)(x=-1+-2sqrt2$

$color(blue)(x^2+2x-7$

This equation is a Quadratic equation (in form $ax^2+bx+c=0$ )

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

In this case $color(red)(a=1,b=2,c=-7$

Substitute the values:

$rarrx=color(orange)((-2+-sqrt(2^2-4(1)(-7)))/(2(1))$

$rarrx=color(orange)((-2+-sqrt(4-(-28)))/(2)$

$rarrx=color(orange)((-2+-sqrt(4+28))/2$

$rarrx=color(orange)((-2+-sqrt32)/2$

$rarrx=color(orange)((-2+-sqrt(16*2))/2$

$rarrx=color(orange)((-2+-4sqrt2)/2$

$rarrx=color(orange)(-2/2+-(4sqrt2)/2$

$rarrx=color(orange)(-cancel(2)/cancel(2)+-(cancel4sqrt2)/cancel2$

$rArrcolor(green)(x=-1+-2sqrt2$

How do you solve x^2 + 2x-7=0x^2 + 2x-7=0 using the quadratic formula?