If it's not what You are looking for type in the equation solver your own equation and let us solve it.
5x^2+26x+7=0
a = 5; b = 26; c = +7;
Δ = b2-4ac
Δ = 262-4·5·7
Δ = 536
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{536}=\sqrt{4*134}=\sqrt{4}*\sqrt{134}=2\sqrt{134}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{134}}{2*5}=\frac{-26-2\sqrt{134}}{10} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{134}}{2*5}=\frac{-26+2\sqrt{134}}{10} $
| x+3=3/4 | | -2(x-7)=-6 | | -32=-8-12x | | x2=164 | | 5m=+1+1+3m+6(2m+4) | | ((x+3)/3)-((4x-1)/6)=3+x/3 | | x-150/50=5 | | -5(1-x)=4(x-1) | | 4x+2+4x+3=45 | | 3y-1+y+49=5y+50-3y | | x/5+8=11 | | 164=x2 | | 14-(2c+5)=(-2)+9 | | 8a-(5a+1)=8-2(1a-3) | | 3y-1+y+49=5y+55-3y | | 2(y-4)=7 | | 2(c+1=10 | | 28=3u-11 | | (10x-7)=0 | | 8(7+a)=160 | | 6a-10+2a-30-50+40a=3+2a-48a+21-(-30) | | 10(1/5x+2)=10(1/5x+1 | | 13v+29=6v-27 | | 4z/9+5=-5 | | 2-4x-8=14 | | 6y-60=-30 | | n2+n-2=60 | | -3(x-5)-10=20 | | 5(1+3b)+8b=-3b+7(4+7b) | | x/5+7/2=-3/2 | | 3x^+9x=0 | | 3(5x+18)=114 |