How do I algebraically solve |4x-3|=5√ (x+4)

Step-by-step explanation:

|4x-3|=5√ (x+4) ⇔ |4x-3|²=5² (√ (x+4)) ² and x+4 ≥ 0

⇔ (4x-3) ² = 25 (x+4) and x+4 ≥ 0 (because : / a/² = a²)

⇔16x²-24x+9 = 25x + 100 and x+4 ≥ 0

⇔ 16x² - 49x - 91 = 0 and x+4 ≥ 0 quadratic equation

Δ = (-49) ²-4 (16) (-91) = 8225

two solution : X1 = (49-√8225) / 32 ≅ - 1.3 accept (-1.3+4 ≥ 0)

X2 = (49+√8225) / 32 ≅4.37 accept (4.37+4 ≥ 0)