Question

Right answers to this easy easy problems gets brainliest & show work if possible

Answer 1

Step-by-step explanation:

For a quadratic in the form of $ax^{2} + bx + c = 0$, the solutions for $x$ can be found using the quadratic formula:

$\frac{-b ± \sqrt{b^{2} - 4ac}}{2a}$

(1) $m^{2} - 5m - 14 = 0$

First, let's get the values for $a, b, c$:

$a = 1$

$b = -5$

$c = -14$

Next, let's plug them into the quadratic equation:

$\frac{-(-5) ± \sqrt{(-5)^{2} - 4(1)(-14)}}{2(1)}$

$\frac{5 ± \sqrt{25 + 56}}{2}$

$\frac{5 ± \sqrt{81}}{2}$

$\frac{5 ± 9}{2}$

With this, we know the two solutions for $m$ are

$\frac{5 + 9}{2} = \frac{14}{2} = 7$

$\frac{5 - 9}{2} = \frac{-4}{2} = -2$

(2) $b^{2} - 4b + 4 = 0$

First, let's get the values for $a, b, c$:

$a = 1$

$b = -4$

$c = 4$

Next, let's plug them into the quadratic equation:

$\frac{-(-4) ± \sqrt{(-4)^{2} - 4(1)(4)}}{2(1)}$

$\frac{4 ± \sqrt{16 - 16}}{2}$

$\frac{4 ± \sqrt{0}}{2}$

$\frac{4}{2}$

With this, we know the only solution for $b$ is $2$.

Answer 2

Answer:

For a quadratic in the form of , the solutions for  can be found using the quadratic formula:

(1)

First, let's get the values for :

Next, let's plug them into the quadratic equation:

With this, we know the two solutions for  are

(2)

First, let's get the values for :

Next, let's plug them into the quadratic equation:

With this, we know the only solution for  is .

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Step-by-step explanation: