Using the quadratic formula, solve 2z² − 6z + 7 = 0. Given that z = -2 + 5i, determine the argument of z, the modulus of z, and (-2 + 5i)⁶ using De Moivre's theorem.

For the quadratic equation $az^2 + bz + c = 0$, the solutions are given by $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Here, $a=2$, $b=-6$, $c=7$.

$\boxed{z = \frac{3}{2} \pm \frac{\sqrt{5}}{2}i}$

For $z = -2 + 5i$, the real part $x = -2$ and the imaginary part $y = 5$. Since $x < 0$ and $y > 0$, $z$ lies in the second quadrant.

The reference angle is $\alpha = \arctan\left(\frac{|y|}{|x|}\right)$.

For a complex number in the second quadrant, the argument is $\theta = 180^\circ - \alpha$.

$\boxed{\arg(z) = 180^\circ - \arctan\left(\frac{5}{2}\right) \approx 111.80^\circ}$

For $z = x + yi$, the modulus is $|z| = \sqrt{x^2 + y^2}$.
Here, $x = -2$ and $y = 5$.

$\boxed{|z| = \sqrt{29}}$

From parts (i) and (ii), we have $r = |z| = \sqrt{29}$ and $\theta = 180^\circ - \arctan\left(\frac{5}{2}\right) \approx 111.8014^\circ$.

De Moivre's theorem states that $(r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))$.
Here, $n=6$.

The angle $6\theta \approx 6 \times 111.8014^\circ = 670.8084^\circ$.
To express the angle in the range $[0^\circ, 360^\circ)$, subtract $360^\circ$:
$670.8084^\circ - 360^\circ = 310.8084^\circ$.

Converting to rectangular form:
$\cos(310.8084^\circ) \approx 0.6535$
$\sin(310.8084^\circ) \approx -0.7569$

$\boxed{(-2+5i)^6 \approx 24389(\cos(310.81^\circ) + i\sin(310.81^\circ)) \approx 15939.8 - 18456.6i}$