Question

4x^2+20x+25=0 then x=?and d= ?​

Answer 1

Answer:

$\bf \huge {4x}^{2} + 20x + 25 = 0$

$\bf \huge x = \frac{ -20 \pm \sqrt{{20}^{2} - 4 \times 4 \times 25}}{2 \times 4 }$

$\bf \huge x = \frac{ -20 \pm \sqrt{400-400}}{2 \times 4}$

$\bf \huge x = \frac{ -20 \pm \sqrt{0}}{8}$

$\bf \huge x = \frac{ -20 \pm 0 }{8}$

$\bf \huge x = \frac{-20}{8}$

$\bf \huge x = \frac{-5}{2}$

$\bf \huge\boxed{ x = -\frac{5}{2}}$