$a\text{)}3x^2+5x=2$  

$3x^2+5x-2=0$  

$\Delta =5^2-4\cdot 3\cdot \left(-2\right)=25+24=49\sqrt{\Delta }=7$  

$x_1=\frac{-5-7}{2\cdot 3}=\frac{-12}{6}=-2$  

$x_2=\frac{-5+7}{2\cdot 3}=\frac{2}{6}=\frac{1}{3}$  

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$b\text{)}5x=2x^2+3$  

$2x^2-5x+3=0$

$\Delta ={\left(-5\right)}^2-4\cdot 2\cdot 3=25-24=1,\sqrt{\Delta }=1$   

$x_1=\frac{5-1}{2\cdot 2}=\frac{4}{4}=1$  

$x_2=\frac{5+1}{2\cdot 2}=\frac{6}{4}=\frac{3}{2}=1\frac{1}{2}$  

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$c\text{)}4x-x^2=7$  

$-x^2+4x-7=0$  

$\Delta =4^2-4\cdot \left(-1\right)\cdot \left(-7\right)=16-28<0$  

To równanie jest sprzeczne.

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$d\text{)}6\left(x^2+2\right)=4\left(3-x\right)$  

$6x^2+12=12-4x$  

$6x^2+4x=0$  

$2x\left(3x+2\right)=0$  

zatem:

$2x=0\vee 3x+2=0$  

$x=0\vee 3x=-2$  

$x=0\vee x=-\frac{2}{3}$  

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$e\text{)}5x^2=2-9x$  

$5x^2+9x-2=0$  

$\Delta =9^2-4\cdot 5\cdot \left(-2\right)=81+40=121,\sqrt{\Delta }=11$  

$x_1=\frac{-9-11}{2\cdot 5}=\frac{-20}{10}=-2$  

$x_2=\frac{-9+11}{2\cdot 5}=\frac{2}{10}=\frac{1}{5}$  

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$f\text{)}{\left(2x+3\right)}^2=8x+8$  

$4x^2+12x+9=8x+8$  

$4x^2+4x+1=0$  

${\left(2x+1\right)}^2=0$  

zatem:

$2x+1=0$  

$2x=-1$  

$x=-\frac{1}{2}$