Question

1. $x^{2}-11x + 19=-5$ 6) $n^{2}+7n + 15 = 5$ 7) $n^{2}-10n + 22=-2$ 8) $n^{2}+3n-12 = 6$ 9) $6n^{2}-18n-18 = 6$ 10) $7r^{2}-14r=-7$

Explanation:

Step1: Rearrange to standard quadratic form

For equation (5) $x^{2}-11x + 19=-5$, rewrite it as $x^{2}-11x+24 = 0$.
For equation (6) $n^{2}+7n + 15 = 5$, rewrite it as $n^{2}+7n + 10=0$.
For equation (7) $n^{2}-10n + 22=-2$, rewrite it as $n^{2}-10n+24 = 0$.
For equation (8) $n^{2}+3n-12 = 6$, rewrite it as $n^{2}+3n - 18=0$.
For equation (9) $6n^{2}-18n-18 = 6$, rewrite it as $6n^{2}-18n-24 = 0$ or $n^{2}-3n - 4=0$ (dividing by 6).
For equation (10) $7r^{2}-14r=-7$, rewrite it as $7r^{2}-14r + 7=0$ or $r^{2}-2r + 1=0$ (dividing by 7).

Step2: Solve quadratic equations

Equation (5):

For $x^{2}-11x + 24=0$, use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ where $a = 1$, $b=-11$, $c = 24$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-11)^{2}-4\times1\times24=121 - 96 = 25$.
Then $x=\frac{11\pm\sqrt{25}}{2}=\frac{11\pm5}{2}$. So $x_1=\frac{11 + 5}{2}=8$ and $x_2=\frac{11-5}{2}=3$.

Equation (6):

For $n^{2}+7n + 10=0$, factor it as $(n + 2)(n+5)=0$. So $n=-2$ or $n=-5$.

Equation (7):

For $n^{2}-10n + 24=0$, factor it as $(n - 4)(n - 6)=0$. So $n = 4$ or $n=6$.

Equation (8):

For $n^{2}+3n-18=0$, factor it as $(n + 6)(n - 3)=0$. So $n=-6$ or $n=3$.

Equation (9):

For $n^{2}-3n - 4=0$, factor it as $(n - 4)(n+1)=0$. So $n = 4$ or $n=-1$.

Equation (10):

For $r^{2}-2r + 1=0$, factor it as $(r - 1)^{2}=0$. So $r = 1$.

Answer:

(5) $x = 3,8$
(6) $n=-2,-5$
(7) $n = 4,6$
(8) $n=-6,3$
(9) $n = 4,-1$
(10) $r = 1$