$$25k^{2}-70k+49-8\left(5k-7\right)+15=0$$
Use the distributive property to multiply $-8$ by $5k-7$.$$25k^{2}-70k+49-40k+56+15=0$$
Combine $-70k$ and $-40k$ to get $-110k$.$$25k^{2}-110k+49+56+15=0$$
Add $49$ and $56$ to get $105$.$$25k^{2}-110k+105+15=0$$
Add $105$ and $15$ to get $120$.$$25k^{2}-110k+120=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $25$ for $a$, $-110$ for $b$, and $120$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.$$k=\frac{-\left(-110\right)±\sqrt{\left(-110\right)^{2}-4\times 25\times 120}}{2\times 25}$$
Square $-110$.$$k=\frac{-\left(-110\right)±\sqrt{12100-4\times 25\times 120}}{2\times 25}$$
Multiply $-4$ times $25$.$$k=\frac{-\left(-110\right)±\sqrt{12100-100\times 120}}{2\times 25}$$
Multiply $-100$ times $120$.$$k=\frac{-\left(-110\right)±\sqrt{12100-12000}}{2\times 25}$$
Add $12100$ to $-12000$.$$k=\frac{-\left(-110\right)±\sqrt{100}}{2\times 25}$$
Take the square root of $100$.$$k=\frac{-\left(-110\right)±10}{2\times 25}$$
The opposite of $-110$ is $110$.$$k=\frac{110±10}{2\times 25}$$
Multiply $2$ times $25$.$$k=\frac{110±10}{50}$$
Now solve the equation $k=\frac{110±10}{50}$ when $±$ is plus. Add $110$ to $10$.$$k=\frac{120}{50}$$
Reduce the fraction $\frac{120}{50}$ to lowest terms by extracting and canceling out $10$.$$k=\frac{12}{5}$$
Now solve the equation $k=\frac{110±10}{50}$ when $±$ is minus. Subtract $10$ from $110$.$$k=\frac{100}{50}$$
Divide $100$ by $50$.$$k=2$$
The equation is now solved.$$k=\frac{12}{5}$$ $$k=2$$