a)
$\,\,\,\,\,\,\,-{{x}^{2}}+3x=2$
$\Leftrightarrow {{x}^{2}}-3x+2=0$
$\Leftrightarrow \left( {{x}^{2}}-3x+\dfrac{9}{4} \right)-\dfrac{1}{4}=0$
$\Leftrightarrow {{\left( x-\dfrac{3}{2} \right)}^{2}}=\dfrac{1}{4}$
$\Leftrightarrow\left[\begin{array}{1}x-\dfrac{3}{2}=\dfrac{1}{2}\\x-\dfrac{3}{2}=-\dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{1}x=2\\x=1\end{array}\right.$
b)
$\,\,\,\,\,\,\,{{x}^{2}}+7x=-10$
$\Leftrightarrow {{x}^{2}}+7x+\dfrac{49}{4}=-10+\dfrac{49}{4}$
$\Leftrightarrow {{\left( x+\dfrac{7}{2} \right)}^{2}}=\dfrac{9}{4}$
$\Leftrightarrow\left[\begin{array}{1}x+\dfrac{7}{2}=\dfrac{3}{2}\\x+\dfrac{7}{2}=-\dfrac{3}{2}\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{1}x=-2\\x=-5\end{array}\right.$
c)
$\,\,\,\,\,\,\,2{{x}^{2}}+3x=5$
$\Leftrightarrow 2{{x}^{2}}+3x+\dfrac{9}{8}=5+\dfrac{9}{8}$
$\Leftrightarrow {{\left( x\sqrt{2}+\dfrac{3\sqrt{2}}{4} \right)}^{2}}=\dfrac{49}{8}$
$\Leftrightarrow\left[\begin{array}{1}x\sqrt{2}+\dfrac{3\sqrt{2}}{4}=\dfrac{7\sqrt{2}}{4}\\x\sqrt{2}+\dfrac{3\sqrt{2}}{4}=-\dfrac{7\sqrt{2}}{4}\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{1}x\sqrt{2}=\sqrt{2}\\x\sqrt{2}=-\dfrac{5\sqrt{2}}{2}\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{1}x=1\\x=-\dfrac{5}{2}\end{array}\right.$
d)
$\,\,\,\,\,\,\,{{x}^{2}}-7x=-12$
$\Leftrightarrow {{x}^{2}}-7x+\dfrac{49}{4}=-12+\dfrac{49}{4}$
$\Leftrightarrow {{\left( x-\dfrac{7}{2} \right)}^{2}}=\dfrac{1}{4}$
$\Leftrightarrow\left[\begin{array}{1}x-\dfrac{7}{2}=\dfrac{1}{2}\\x-\dfrac{7}{2}=-\dfrac{1}{2}\end{array}\right.$
$\Leftrightarrow\left[\begin{array}{1}x=4\\x=3\end{array}\right.$
e)
$\,\,\,\,\,\,\,9{{x}^{2}}+6x=-1$
$\Leftrightarrow 9{{x}^{2}}+6x+1=-1+1$
$\Leftrightarrow {{\left( 3x+1 \right)}^{2}}=0$
$\Leftrightarrow 3x+1=0$
$\Leftrightarrow x=-\dfrac{1}{3}$
