Помогите пожалуйста сделать,через дискриминант ,которые обведены,пожалуйста

Question 1

Question 2

434) $18-x^2=14\\18-14=x^2\\x^2=4\\x_1=-2;x_2=2$

436) $x^{2} +2=x+2;\\x^2-x+2-2=0$
$x^{2} -x=0$
D=1-4*1*0=1
$x_1= \frac{1-1}{2} =0$ или $x_2= \frac{1+1}{2} =1$

438) $2x^2+3=3-7x\\2x^2+7x+3-3=0\\2x^2+7x=0$
D=49-4*2*0=49=7^2
$x_1= \frac{-7-7}{4} =- \frac{14}{4} =-3,5\\x_2= \frac{-7+7}{4}=0$

440) $x^{2} +2x=16x-49;\\x^2-14x+49=0\\D=256-4*49=0\\x= \frac{14}{2} =7$

442) $5x^2+1=6x-4x^2;\\9x^2-6x+1=0\\D=36-4*9=0\\x= \frac{6}{2*9} = \frac{1}{3}$

444) x(x+3)=4
$x^{2} +3x-4=0\\D=9-4*(-4)=25=5^2\\ x_1= \frac{-3-5}{2} =-4;\\x_2= \frac{-3+5}{2} =1.$

446) x(x-4)=-3
$x^{2} -4x+3=0\\D=16-4*3=4=2^2;\\x_1= \frac{4-2}{2}=1;\\x_2=\frac{4+2}{2}=3.$

448) x(2x-3)=4x-3;
$2x^2-7x+3=0\\D=49-4*2*3=25=5^2;\\x_1= \frac{7-5}{2*2}= \frac{2}{4}=0,5\\x_2= \frac{7+5}{2*2}= \frac{12}{4}=3$

450) (3x+1)(6-4x)=0;
$18x+6-12x^2-4x=0;\\-12x^2+14x+6=0\\D=196-4*6*(-12)=484=22^2\\x_1=- \frac{1}{3} ;\\x_2= \frac{6}{4}=1,5$

452) (6x+3)(9-x)=0;
$-6x^2+51x+27=0;\\D=2601-4*27*(-6)=3249=57^2\\x_1= \frac{-51-57}{2*(-6)}= \frac{108}{12} =9;\\x_2= \frac{-51+57}{2*(-6)}=- \frac{6}{12} =-0,5.$

454) (x-1)(5x+0,5)=0;
$5 x^{2} -4,5x-0,5=0\\D=20,25-4*5*(-0,5)=30,25=5,5^2;\\x_1= \frac{4,5-5,5}{10} =-0,1\\x_2= \frac{4,5+5,5}{10}=1$

456) 2(5x-7)(1+x)=0
$2(5x^2-2x-7)=0;\\5x^2-2x-7=0;\\D=4-4*(-7)*5=144=12^2\\x_1= \frac{2-12}{10}=-1;\\x_2= \frac{2+12}{10}=1,4$

458) (6-x)(5x+40)=0;
$-5x^2-10x+240=0\\D=100-4*(-5)*240=4900=70^2\\x_1= \frac{10+70}{2*(-5)}=-8;\\x_2= \frac{10-70}{2*(-5)}= \frac{60}{10} =6$