Question

Solve for the roots in simplest form using the quadratic formula:4x2 – 12x = -5​

Answer 1

Answer:

x=2.5&.5

Step-by-step explanation:

The quadratic formula is (-b+or-sqrt(b^2-4ac)/2a

12+or-sqrt((-12^2)-4(4)(5))

12+or-sqrt(144-80)

12+or-sqrt(64)

(12+8)/8 and (12-8)/8

x=2.5 and x=.5

Answer 2

Answer: $x=\frac{5}{2},\:x=\frac{1}{2}$ or $x=2.5,\:x=0.5$

Step-by-step explanation:

$4x^2-12x=-5$

$\mathrm{Add\:}5\mathrm{\:to\:both\:sides}$

$4x^2-12x+5=-5+5$

$\mathrm{Simplify}$

$4x^2-12x+5=0$

Solve with Quadratic Formula

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$a=4,\:b=-12,\:c=5:\quad x_{1,\:2}=\frac{-\left(-12\right)\pm \sqrt{\left(-12\right)^2-4\cdot \:4\cdot \:5}}{2\cdot \:4}$

$\frac{-\left(-12\right)+\sqrt{\left(-12\right)^2-4\cdot \:4\cdot \:5}}{2\cdot \:4}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a$

$\frac{12+\sqrt{\left(-12\right)^2-4\cdot \:4\cdot \:5}}{2\cdot \:4}$

$12+\sqrt{\left(-12\right)^2-4\cdot \:4\cdot \:5}=12+\sqrt{64}$

$\sqrt{\left(-12\right)^2-4\cdot \:4\cdot \:5}=\sqrt{64}$

$\sqrt{64}=8$

$=\frac{12+8}{8}$

$=\frac{20}{8}$

$=\frac{5}{2}$

$x=\frac{5}{2},\:x=\frac{1}{2}$