M_POQ = (3x +7) mZQOR = (5x - 2)° mZPOR = 61°

Step-by-step explanation:

3x+7)(5x-2)=61

We move all terms to the left:

(3x+7)(5x-2)-(61)=0

We multiply parentheses ..

(+15x^2-6x+35x-14)-61=0

We get rid of parentheses

15x^2-6x+35x-14-61=0

We add all the numbers together, and all the variables

15x^2+29x-75=0

a = 15; b = 29; c = -75;

Δ = b2-4ac

Δ = 292-4·15·(-75)

Δ = 5341

The delta value is higher than zero, so the equation has two solutions

We use following formulas to calculate our solutions:

x1=−b−Δ√2ax2=−b+Δ√2a

The end solution:

Δ−−√=5341−−−−√=49∗109−−−−−−−√=49−−√∗109−−−√=7109−−−√

x1=−b−Δ√2a=−(29)−7109√2∗15=−29−7109√30

x2=−b+Δ√2a=−(29)+7109√2∗15=−29+7109√30

r and t.

both are 64 degrees.