Question

Please help!!!!! 

Answer 1

Part 1) Use the quadratic formula to solve the equation.

4x^2−10x+5=0

we know that
the quadratic formula is
$x=[-b (+/-) \sqrt{b^{2} -4ac}]/ 2a$
in this problem
a=4
b=-10
c=5

$x=[10 (+/-) \sqrt{10^{2} -4*(4)*(5)}]/(2*4)\\ x=[10(+/-) \sqrt{20}]/8 \\ x1=[10+ \sqrt{20}]/8 \\ x2=(10- \sqrt{20}]/8$

the answer Part 1) is

$x1=(5+ \sqrt{5})/4 \\ x2=(5- \sqrt{5} )/4$

Part 2)  What are the zeros of the function f(x) =x^2+7x−18?

we have 
f(x)=x²+7x-18
equals to zero the function
x²+7x-18=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

x²+7x=18

Complete the square. Remember to balance the equation by adding the same constants to each side
x²+7x+12.25=18+12.25

Rewrite as perfect squares

(x+3.5)²=30.25-----> square root----> (+/-)[x+3.5]=5.5

(+)[x+3.5]=5.5---> x=5.5-3.5----> x=2

(-)[x+3.5]=5.5----> x=-9

the answer part 2) is

the zero of the function are

x=2 and x=-9

Part 3) question of the picture

$(2 x^{2} +4x-8)-(4 x^{2} -x+1) \\ (2 x^{2} -4 x^{2} )+(4x+x)+(-8-1) \\ -2 x^{2} +5x-9 \\ standart form \\ a x^{2} +bx+c=0 \\ -2 x^{2} +5x-9=0$