find the coordinates of the points of intersaction of the straight line of the staight line5x-4y+25=0 and the circle 2x^2+2y^2+7x-6Y-15=0.do it with full calculations

Sure, let's solve this step by step.

First, let's rewrite the equation of the line in the form y = mx + c. The equation 5x - 4y + 25 = 0 can be rewritten as y = (5/4)x + 25/4.

Next, let's rewrite the equation of the circle in the standard form (x - h)^2 + (y - k)^2 = r^2. The equation 2x^2 + 2y^2 + 7x - 6y - 15 = 0 can be rewritten as (x + 7/4)^2 + (y - 3/2)^2 = 169/16.

Now, we substitute y from the equation of the line into the equation of the circle:

(x + 7/4)^2 + ((5/4)x + 25/4 - 3/2)^2 = 169/16.

Solving this equation will give us the x-coordinates of the points of intersection. Let's simplify the equation:

(x^2 + 7x/2 + 49/16) + ((25x/16 + 125/16 - 6/4)^2) = 169/16.

This simplifies to:

x^2 + 7x/2 + 25x/16 + 49/16 + 125/16 - 3/2 = 169/16.

Combining like terms gives:

41x/16 + 98/16 = 169/16.

Solving for x gives x = 71/41 and x = -98/41.

Substituting these x-values into the equation of the line will give us the corresponding y-values:

For x = 71/41, y = (5/4)(71/41) + 25/4 = 355/164 + 100/164 = 455/164.

For x = -98/41, y = (5/4)(-98/41) + 25/4 = -490/164 + 100/164 = -390/164.

So, the points of intersection are (71/41, 455/164) and (-98/41, -390/164).