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

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Answer 1 · 2016-02-14T09:41:52

$x = - 1 \pm 2 \sqrt{2}$ $color(green)(x=-1+-2sqrt2$

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

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

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(brown)(x=(-b+-sqrt(b^2-4ac))/(2a)$

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

Substitute the values:

$\to x = \frac{- 2 \pm \sqrt{2^{2} - 4 (1) (- 7)}}{2 (1)}$ $rarrx=color(orange)((-2+-sqrt(2^2-4(1)(-7)))/(2(1))$

$\to x = \frac{- 2 \pm \sqrt{4 - (- 28)}}{2}$ $rarrx=color(orange)((-2+-sqrt(4-(-28)))/(2)$

$\to x = \frac{- 2 \pm \sqrt{4 + 28}}{2}$ $rarrx=color(orange)((-2+-sqrt(4+28))/2$

$\to x = \frac{- 2 \pm \sqrt{32}}{2}$ $rarrx=color(orange)((-2+-sqrt32)/2$

$\to x = \frac{- 2 \pm \sqrt{16 \cdot 2}}{2}$ $rarrx=color(orange)((-2+-sqrt(16*2))/2$

$\to x = \frac{- 2 \pm 4 \sqrt{2}}{2}$ $rarrx=color(orange)((-2+-4sqrt2)/2$

$\to x = - \frac{2}{2} \pm \frac{4 \sqrt{2}}{2}$ $rarrx=color(orange)(-2/2+-(4sqrt2)/2$

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

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

How do you solve x^2 + 2x-7=0x2+2x−7=0x^2 + 2x-7=0 using the quadratic formula?