$$ b_{1}(t) :=3+t-2 t^{2}, b_{2}(t) :=-5-5 t+3 t^{2} \text { und } b_{3}(t) :=3-2 t^{2} $$ Want to show that this is a basis so linear independent
$$\begin{pmatrix} -2a & 3b &-2c \\ a & -5b& 0 \\3a&-5b&3c \end{pmatrix}\begin{pmatrix} t^2\\t\\1 \end{pmatrix}$$ = 0 for each row is this a correct way?
Sort of. There's absolutely no call for putting polynomials in matrices, and the fact that you have suggests a (very common) gap in your understanding. It would be better if you worked from definitions, instead of guessing how best to put things in matrices.
To investigate the linear independence of $b_1, b_2, b_3$, we must solve $$\lambda_1 b_1 + \lambda_2 b_2 + \lambda_3 b_3 = 0.$$ You should be aware that $0$ is a polynomial (the function that always returns $0$), the $+$ operation is pointwise addition between polynomials, and the multiplication here is scalar multiplication between scalars and polynomials. According to the definitions of addition and scalar multiplication of polynomials, this is equivalent to $$\lambda_1 b_1(t) + \lambda_2 b_2(t) + \lambda_3 b_3(t) = 0(t) = 0$$ for all $t$. Or equivalently, for all $t$, $$\lambda_1(3+t−2t^2)+ \lambda_2(−5−5t+3t^2) + \lambda_3(3−2t^2) = 0.$$ From here, you can proceed a few different ways. The one that would get you closest to your matrix would be to equate coefficients (the fact you can do this is actually a consequence of the linear independence of the monomials $1, x, x^2, \ldots$). If we collect like terms, $$(-2\lambda_1 + 3\lambda_2 - 2 \lambda_3)t^2 + (\lambda_1 - 5\lambda_2)t + (3\lambda_1 - 5 \lambda_2 + 3\lambda_3) = 0t^2 + 0t + 0$$ for all $t$, we can then equate the coefficients to obtain a system of linear equations: \begin{align*} -2\lambda_1 + 3\lambda_2 - 2 \lambda_3 &= 0 \\ \lambda_1 - 5\lambda_2 &= 0 \\ 3\lambda_1 - 5 \lambda_2 + 3\lambda_3 &= 0. \end{align*} This equation can be solved using an augmented matrix $$\left(\begin{array}{ccc|c}-2 & 3 & -2 & 0 \\ 1 & -5 & 0 & 0 \\ 3 & -5 & 3 & 0\end{array}\right),$$ which can be solved in the usual way.
Please note that the coefficients of $b_1, b_2, b_3$ did not make it into the matrix, nor did the variable $t$ or any of its powers. When equating coefficients, take only the coefficients!