Odpowiedź :
to jest na 1 zadanie dalej wystarczy zrobić tak samo
1)
[tex]( x - 4 )^2 = 17 - 9x\\x^2-8x+16=17-9x\\x^2+x-1=0\\\Delta=b^2-4ac=1^2-4*1*\left(-1\right)=1+4=5\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{-1-\sqrt{5}}{2*1}=\cfrac{-1-\sqrt{5}}{2}\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{-1+\sqrt{5}}{2*1}=\cfrac{-1+\sqrt{5}}{2}[/tex]
2)
[tex](x-1)^2=36\\x^2-2x+1=36\\x^2-2x-35=0\\\Delta=b^2-4ac=\left(-2\right)^2-4*1*\left(-35\right)=4+140=144\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{2-\sqrt{144}}{2*1}=\cfrac{2-12}{2}=-5\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{2+\sqrt{144}}{2*1}=\cfrac{2+12}{2}=7[/tex]
3)
[tex]x ( x - 2 ) + 3 ( x - 2 ) = 0\\x^2-2x+3x-6=0\\x^2+x-6=0\\\Delta=b^2-4ac=1^2-4*1*\left(-6\right)=1+24=25\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{-1-\sqrt{25}}{2*1}=\cfrac{-1-5}{2}=-3\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{-1+\sqrt{25}}{2*1}=\cfrac{-1+5}{2}=2[/tex]
4)
[tex]5 ( 2x + 3 ) - 4 x ( 2x + 3 ) = 0\\10x+15-8x^2-12x=0\\-8x^2-2x+15=0\\\Delta=b^2-4ac=\left(-2\right)^2-4*\left(-8\right)*15=4+480=484\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{2-\sqrt{484}}{2*\left(-8\right)}=\cfrac{2-22}{\left(-16\right)}=1.25\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{2+\sqrt{484}}{2*\left(-8\right)}=\cfrac{2+22}{\left(-16\right)}=-1.5[/tex]
5)
[tex]7 ( x + 5 ) = x ( x + 5 )\\7x+35=x^2+5x\\-x^2+2x+35=0\\\Delta=b^2-4ac=2^2-4*\left(-1\right)*35=4+140=144\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{-2-\sqrt{144}}{2*\left(-1\right)}=\cfrac{-2-12}{\left(-2\right)}=7\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{-2+\sqrt{144}}{2*\left(-1\right)}=\cfrac{-2+12}{\left(-2\right)}=-5[/tex]
6)
[tex]( 7 - x )^2 = 4x^2\\49-14x+x^2=4x^2\\-3x^2-14x+49=0\\\Delta=b^2-4ac=\left(-14\right)^2-4*\left(-3\right)*49=196+588=784\\x_1=\cfrac{-b-\sqrt\Delta}{2a}=\cfrac{14-\sqrt{784}}{2*\left(-3\right)}=\cfrac{14-28}{\left(-6\right)}=2\frac{1}{3}\\x_2=\cfrac{-b+\sqrt\Delta}{2a}=\cfrac{14+\sqrt{784}}{2*\left(-3\right)}=\cfrac{14+28}{\left(-6\right)}=-7[/tex]