How do you solve $x^{2} - 4 x = 5$ $x^2 - 4x = 5$ using the quadratic formula?

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How do you solve $x^{2} - 4 x = 5$ $x^2 - 4x = 5$ using the quadratic formula?

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Answer 1 · 2017-06-01T13:09:18

$x = - 1 or x = 5$ $color(blue)(x=-1 or x=5$

Explanation:

General quadratic equation:

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

Quadratic formula:

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

$x^{2} - 4 x = 5$ $x^2-4x=5$

$:.x^2-4x-5=0$

$:.a=1,b=-4,c=-5$

$:.x=(-(-4)+-sqrt((-4)^2-4(1)(-5)))/(2(1))$

$:.x=(4+-sqrt(16+20))/2$

$:.x=(4+-sqrt36)/2$

$:.x=(4+-6)/2$

$:.x=(4+6)/2$

or:

$:.x=(4-6)/2$

$:.x=10/2$ or $x=(-2)/2$

$:.color(blue)(x=5$ or $color(blue)(x=-1$

substitute $color(blue)(x=-1$

$:.(-1)^2-4(-1)=5$

$:.1+4=5$

$:.color(blue)(color(blue)(5=5$

substitute $color(blue)(x=5$

$:.(5)^2-4(5)=5$

$:.25-20=5$

$:.color(blue)(5=5$

Answer 2 · 2017-06-01T13:12:07

$x=5, x=-1$

Explanation:

Subtract $5$ from both sides.

$x^2-4x-5=0$

From this, $a=1, b=-4, c=-5$

The quadratic formula is $(-b+-sqrt(b^2-4ac))/(2a)$

Substituting the values

$(-(-4)+-sqrt((-4)^2-4*1*-5))/(2*1)$

$(4+-sqrt(16+20))/2$

$(4+sqrt(36))/2$

$4/2+sqrt(36)/2, 4/2-sqrt(36)/2$

$2+sqrt(36)/2, 2-sqrt(36)/2$

$2+6/2, 2-6/2$

$2+3, 2-3$

$5, -1$

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How do you solve $x^{2} - 4 x = 5$ $x^2 - 4x = 5$ using the quadratic formula?

2 Answers

Explanation:

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How do you solve $x^{2} - 4 x = 5$ $x^2 - 4x = 5$ using the quadratic formula?