Is someone able to put this equation in quadratic form? 4x+5=8√(1-x^2 )?

First, square both sides:
∴ (4x + 5)^2 = (8√(1-x^2))^2
∴ 16x^2 + 40x + 25 = 64 (1-x^2)
∴ 16x^2 +40x + 25 = 64 - 64x^2
Now rearrange to get the required quadratic equation;
∴ 80x^2 + 40x - 39 = 0
Hope this helps !!!!!!!!

4x + 5 = 8√(1 - x² )

[4x + 5]² = [8√(1 - x² )]²

16x² + 40x + 25 = 64(1 - x²)

16x² + 40x + 25 = 64 - 64x²

80x² + 40x - 39 = 0

x = {-40 ± √[40² - 4(80)(-39)]} / [2(80)]

x = (-5 + 2√55)/20 or x = (-5 - 2√55)/20

First square both sides getting 16x^2 + 40x + 25 = 64 - 64x^2. Then 80x^2 + 40x - 39 = 0.