How do you solve $4r ^ { 2} + 4r - 8= - 5$ ?

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Answer 1 · 2018-03-10T02:14:16

$r=-3/2,1/2$

Solve:

$4r^2+4r-8=-5$

Add $5$ to both sides.

$4r^2+4r-8+5=0$

Simplify.

$4r^2+4r-3=0$ is a quadratic equation in standard form:

$ax^2+bx+c=0,$

where:

$a=4,$ $b=4,$ $c=-3$

You can solve this equation by using the quadratic formula.

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

Substitute $r$ for $x$ , plug in the known values and solve.

$r=(-4+-sqrt(4^2-4*4*-3))/(2*4)$

$r=(-4+-sqrt(64))/8$

Simplify $sqrt64$ to $8$ .

$r=(-4+-8)/8$

$r=(-4+8)/8,$ $(-4-8)/8$

$r=4/8,$ $-12/8$

Reduce.

$r=1/2,$ $-3/2$

How do you solve 4r ^ { 2} + 4r - 8= - 54r ^ { 2} + 4r - 8= - 5?