$\text{a)}12x^2+4=40$  

$12x^2=36\mid :12$  

$x^2=3$  

$x=\sqrt{3}\text{lub}x=-\sqrt{3}$  

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$\text{b)}\frac{3x^2}{2}-5=0$  

$\frac{3}{2}{x}^2=5\mid \cdot \frac{2}{3}$  

$x^2=\frac{10}{3}$  

$x=\frac{\sqrt{10}}{\sqrt{3}}=\frac{\sqrt{30}}{3}\text{lub}x=-\frac{\sqrt{10}}{3}=-\frac{\sqrt{30}}{3}$  

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$\text{c)}2x^2-7x=0\mid :2$  

$x^2-\frac{7}{2}x=0$  

$x\left(x-\frac{7}{2}\right)=0$  

$x=0\text{lub}x-\frac{7}{2}=0$  

$x=0\text{lub}x=\frac{7}{2}$  

$x=0\text{lub}x=3\frac{1}{2}$  

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$\text{d)}\frac{1}{5}{x}^2=3x$  

$\frac{1}{5}{x}^2-3x=0\mid \cdot 5$  

$x^2-15x=0$  

$x\left(x-15\right)=0$  

$x=0\text{lub}x-15=0$  

$x=0\text{lub}x=15$  

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$\text{e)}6x^2+5x=3x^2-x$  

$3x^2+6x=0\mid :3$  

$x^2+2x=0$  

$x\left(x+2\right)=0$  

$x=0\text{lub}x+2=0$  

$x=0\text{lub}x=-2$  

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$\text{f)}\frac{1}{2}{x}^2+5=3-\frac{1}{4}{x}^2$  

$\frac{2}{4}{x}^2+\frac{1}{4}{x}^2=-2$  

$\frac{3}{4}{x}^2=-2\mid \cdot \frac{4}{3}$  

$x^2=-\frac{8}{3}<0$  

Równanie jest sprzeczne (nie ma rozwiązań), ponieważ kwadrat dowolnej liczby rzeczywistej jest nieujemny.

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$\text{g)}{\left(2x-1\right)}^2=25$  

$2x-1=5\text{lub}2x-1=-5$  

$2x=6\text{lub}2x=-4\mid :2$  

$x=3\text{lub}x=-2$  

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$\text{h)}3{\left(x+1\right)}^2-2=0$  

$3{\left(x+1\right)}^2=2\mid :3$  

${\left(x+1\right)}^2=\frac{2}{3}$  

$x+1=\frac{\sqrt{2}}{\sqrt{3}}\text{lub}x+1=-\frac{\sqrt{2}}{\sqrt{3}}$  

$x+1=\frac{\sqrt{6}}{3}\text{lub}x+1=-\frac{\sqrt{6}}{3}$  

$x=\frac{\sqrt{6}}{3}-1\text{lub}x=-\frac{\sqrt{6}}{3}-1$